##### Class 12^{th} Chemistry Part I CBSE Solution

**Intext Questions Pg-68**- How would you determine the standard electrode potential of the system Mg2+|Mg?…
- Can you store copper sulphate solutions in a zinc pot?
- Consult the table of standard electrode potentials and suggest three substances that can…

**Intext Questions Pg-73**- Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.…
- Calculate the emf of the cell in which the following reaction takes place: Ni(S) + 2Ag+…
- The Cell in which the following reaction occurs: 2Fe3+(aq) + 2I- (aq) → 2Fe2+(aq) + I2(s)…

**Intext Questions Pg-83**- Why does the conductivity of a solution decrease with dilution?
- Suggest a way to determine the Î›°m value of water
- The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm^2 mol-1. Calculate its…

**Intext Questions Pg-86**- If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many…
- Suggest a list of metals that are extracted electrolytically.
- Consider the reaction: Cr2O72- + 14H+ + 6e-⇒ 2Cr3+ + 7H2O. What is the quantity of…

**Intext Questions Pg-90**- Write the chemistry of recharging the lead storage battery, highlighting all the materials…
- Suggest two materials other than hydrogen that can be used as fuels in fuel cells.…
- Explain how rusting of iron is envisaged as setting up of an electrochemical cell.…

**Exercises**- Arrange the following metals in the order in which they displace each other from the…
- Given the standard electrode potentials, K+ /K = -2.93V, Ag+ /Ag = 0.80V, Hg2+ /Hg = 0.79V…
- Depict the galvanic cell in which the reaction Zn(s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag(s)…
- Calculate the standard cell potentials of galvanic cell in which the following reactions…
- Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg2+…
- In the button cells widely used in watches and other devices the following reaction takes…
- Define conductivity and molar conductivity of the solution of an electrolyte. Discuss…
- The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar…
- The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Î©.…
- The conductivity of sodium chloride at 298 K has been determined at different…
- Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar…
- How much charge is required for the following reductions: (i) 1 mol of Al3 + to Al? (ii) 1…
- How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from…
- How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2?…
- A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5…
- Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4,…
- Using the standard electrode potentials given in Table 3.1, predict if the reaction…
- Predict the products of electrolysis in each of the following: (i) An aqueous solution of…

**Intext Questions Pg-68**

- How would you determine the standard electrode potential of the system Mg2+|Mg?…
- Can you store copper sulphate solutions in a zinc pot?
- Consult the table of standard electrode potentials and suggest three substances that can…

**Intext Questions Pg-73**

- Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.…
- Calculate the emf of the cell in which the following reaction takes place: Ni(S) + 2Ag+…
- The Cell in which the following reaction occurs: 2Fe3+(aq) + 2I- (aq) → 2Fe2+(aq) + I2(s)…

**Intext Questions Pg-83**

- Why does the conductivity of a solution decrease with dilution?
- Suggest a way to determine the Î›°m value of water
- The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm^2 mol-1. Calculate its…

**Intext Questions Pg-86**

- If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many…
- Suggest a list of metals that are extracted electrolytically.
- Consider the reaction: Cr2O72- + 14H+ + 6e-⇒ 2Cr3+ + 7H2O. What is the quantity of…

**Intext Questions Pg-90**

- Write the chemistry of recharging the lead storage battery, highlighting all the materials…
- Suggest two materials other than hydrogen that can be used as fuels in fuel cells.…
- Explain how rusting of iron is envisaged as setting up of an electrochemical cell.…

**Exercises**

- Arrange the following metals in the order in which they displace each other from the…
- Given the standard electrode potentials, K+ /K = -2.93V, Ag+ /Ag = 0.80V, Hg2+ /Hg = 0.79V…
- Depict the galvanic cell in which the reaction Zn(s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag(s)…
- Calculate the standard cell potentials of galvanic cell in which the following reactions…
- Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg2+…
- In the button cells widely used in watches and other devices the following reaction takes…
- Define conductivity and molar conductivity of the solution of an electrolyte. Discuss…
- The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar…
- The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Î©.…
- The conductivity of sodium chloride at 298 K has been determined at different…
- Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar…
- How much charge is required for the following reductions: (i) 1 mol of Al3 + to Al? (ii) 1…
- How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from…
- How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2?…
- A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5…
- Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4,…
- Using the standard electrode potentials given in Table 3.1, predict if the reaction…
- Predict the products of electrolysis in each of the following: (i) An aqueous solution of…

###### Intext Questions Pg-68

**Question 1.**How would you determine the standard electrode potential of the system Mg^{2+}|Mg?

**Answer:**To determine the standard electrode potential of the system Mg^{2+}|Mg, connect it to the standard hydrogen electrode (SHE). Keep the Mg^{2+}|Mg system as cathode and SHE as anode. This is represented as shown below.

Pt(s) | H_{2}(g, 1 bar)| H^{+} (aq, 1 M) ||Mg^{2+} (aq, 1M)| Mg

The electrode potential of a cell is given by

E^{⊖} = E^{⊖}_{R} – E^{⊖}_{L}

Where,

E^{⊖}_{R}- Potential of the half-cell in the right side of the above representation

E^{⊖}_{L}- Potential of the half-cell in the left side of the above representation

It is to be noted that the potential of the standard hydrogen electrode is zero.

Therefore, E^{⊖}_{L} = 0

E^{⊖} = E^{⊖}_{R} – 0

⇒ E^{⊖} = E^{⊖}_{R}

**Question 2.**Can you store copper sulphate solutions in a zinc pot?

**Answer:**__NO__, because Zn is very reactive with Cu. __It reacts with copper sulphate to form zinc sulphate__ i.e., Zn displaces Cu and metallic Cu is also formed.

The reaction is given as:

Zn + CuSO_{4} ⇒ ZnSO_{4} + Cu

**Question 3.**Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions

**Answer:**For a substance to oxidise Fe^{2+}to Fe^{3+} ion, it must have high reduction potential than Fe^{3+}. The reduction potential of Fe^{3+} to Fe^{2+} reaction is 0.77V, the substances which have reduction potentials higher than this value will oxidise Fe^{2+} ions. Comparing the values, from the table:

F_{2},Br_{2},Cl_{2},H_{2}O_{2} can oxidise ferrous ions.

**Question 1.**

How would you determine the standard electrode potential of the system Mg^{2+}|Mg?

**Answer:**

To determine the standard electrode potential of the system Mg^{2+}|Mg, connect it to the standard hydrogen electrode (SHE). Keep the Mg^{2+}|Mg system as cathode and SHE as anode. This is represented as shown below.

Pt(s) | H_{2}(g, 1 bar)| H^{+} (aq, 1 M) ||Mg^{2+} (aq, 1M)| Mg

The electrode potential of a cell is given by

E^{⊖} = E^{⊖}_{R} – E^{⊖}_{L}

Where,

E^{⊖}_{R}- Potential of the half-cell in the right side of the above representation

E^{⊖}_{L}- Potential of the half-cell in the left side of the above representation

It is to be noted that the potential of the standard hydrogen electrode is zero.

Therefore, E^{⊖}_{L} = 0

E^{⊖} = E^{⊖}_{R} – 0

⇒ E^{⊖} = E^{⊖}_{R}

**Question 2.**

Can you store copper sulphate solutions in a zinc pot?

**Answer:**

__NO__, because Zn is very reactive with Cu. __It reacts with copper sulphate to form zinc sulphate__ i.e., Zn displaces Cu and metallic Cu is also formed.

The reaction is given as:

Zn + CuSO_{4} ⇒ ZnSO_{4} + Cu

**Question 3.**

Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions

**Answer:**

For a substance to oxidise Fe^{2+}to Fe^{3+} ion, it must have high reduction potential than Fe^{3+}. The reduction potential of Fe^{3+} to Fe^{2+} reaction is 0.77V, the substances which have reduction potentials higher than this value will oxidise Fe^{2+} ions. Comparing the values, from the table:

F_{2},Br_{2},Cl_{2},H_{2}O_{2} can oxidise ferrous ions.

###### Intext Questions Pg-73

**Question 1.**Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

**Answer:**Given:

For hydrogen electrode, pH = 10

n = 1

(n = moles of e^{-} from balanced redox reaction)

pH= 10

⇒ [H^{+}] = 10^{-10} M

We know,

Here, E^{0} = 0 (because it is a concentration cell)

P[H_{2}] = 1

⇒

⇒

⇒

⇒

⇒

**Question 2.**Calculate the emf of the cell in which the following reaction takes place:

Ni(S) + 2Ag^{+} (0.002M) → Ni^{2+} (0.160 M) + 2Ag(s)

Given that

**Answer:**Given:

[Ag^{+}] = 0.002 M

[Ni^{2+}] = 0.160 M

n = 2

(n = moles of e^{-} from balanced redox reaction)

E^{0}_{cell}= 1.05 V

Now, using the Nernst equation we get,

⇒

⇒

⇒

**Question 3.**The Cell in which the following reaction occurs:

2Fe^{3+}(aq) + 2I^{–} (aq) → 2Fe^{2+}(aq) + I_{2}(s) has E^{0}_{cell} = 0.236V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

**Answer:**Given:

2Fe^{3+}(aq) + 2I^{–} (aq) → 2Fe^{2+}(aq) + I_{2}(s)

E^{0}_{cell} = 0.236V

n = moles of e^{-} from balanced redox reaction = 2

F = Faraday's constant = 96,485 C/mol

T = 298 K.

Using the formula, we get,

∆_{r}G^{0} = – nFE^{0}_{cell}

⇒ ∆_{r}G^{0} = – 2 × FE^{0}_{cell}

⇒ ∆_{r}G^{0} = −2 × 96485 C mol^{-1} × 0.236 V

⇒ ∆_{r}G^{0} = −45540 J mol^{−1}

⇒ ∆_{r}G^{0} = −45.54 kJ mol^{−1}

Now,

∆_{r}G^{0} = −2.303RT log K_{c}

Where, K is the equilibrium constant of the reaction.

R is the gas constant; R = 8.314 J-mol-C^{-1}

⇒ −45540 J mol^{−1} = –2.303× (8.314 J-mol-C^{-1})× (298 K) × (log Kc)

Solving for K_{c} we get,

⇒ log Kc = 7.98

Taking antilog both side, we get

⇒ Kc = Antilog (7.98)

⇒ K_{c} = 9.6 × 107

__Trick to remember:__

**Question 1.**

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

**Answer:**

Given:

For hydrogen electrode, pH = 10

n = 1

(n = moles of e^{-} from balanced redox reaction)

pH= 10

⇒ [H^{+}] = 10^{-10} M

We know,

Here, E^{0} = 0 (because it is a concentration cell)

P[H_{2}] = 1

⇒

⇒

⇒

⇒

⇒

**Question 2.**

Calculate the emf of the cell in which the following reaction takes place:

Ni(S) + 2Ag^{+} (0.002M) → Ni^{2+} (0.160 M) + 2Ag(s)

Given that

**Answer:**

Given:

[Ag^{+}] = 0.002 M

[Ni^{2+}] = 0.160 M

n = 2

(n = moles of e^{-} from balanced redox reaction)

E^{0}_{cell}= 1.05 V

Now, using the Nernst equation we get,

⇒

⇒

⇒

**Question 3.**

The Cell in which the following reaction occurs:

2Fe^{3+}(aq) + 2I^{–} (aq) → 2Fe^{2+}(aq) + I_{2}(s) has E^{0}_{cell} = 0.236V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

**Answer:**

Given:

2Fe^{3+}(aq) + 2I^{–} (aq) → 2Fe^{2+}(aq) + I_{2}(s)

E^{0}_{cell} = 0.236V

n = moles of e^{-} from balanced redox reaction = 2

F = Faraday's constant = 96,485 C/mol

T = 298 K.

Using the formula, we get,

∆_{r}G^{0} = – nFE^{0}_{cell}

⇒ ∆_{r}G^{0} = – 2 × FE^{0}_{cell}

⇒ ∆_{r}G^{0} = −2 × 96485 C mol^{-1} × 0.236 V

⇒ ∆_{r}G^{0} = −45540 J mol^{−1}

⇒ ∆_{r}G^{0} = −45.54 kJ mol^{−1}

Now,

∆_{r}G^{0} = −2.303RT log K_{c}

Where, K is the equilibrium constant of the reaction.

R is the gas constant; R = 8.314 J-mol-C^{-1}

⇒ −45540 J mol^{−1} = –2.303× (8.314 J-mol-C^{-1})× (298 K) × (log Kc)

Solving for K_{c} we get,

⇒ log Kc = 7.98

Taking antilog both side, we get

⇒ Kc = Antilog (7.98)

⇒ K_{c} = 9.6 × 107

__Trick to remember:__

###### Intext Questions Pg-83

**Question 1.**Why does the conductivity of a solution decrease with dilution?

**Answer:**The conductivity of a solution depends on the amount of ions present per volume of the solution. When diluted, the concentration of the ions decreases which implies that the number of ions per volume decreases thus, in turn, conductivity decreases.

**Question 2.**Suggest a way to determine the Î›°_{m} value of water

**Answer:**Î›°_{m (H2O)} = Î›°_{m(H+)} + Î›°_{m(OH-)}

For calculation purpose, Î›°_{m(Na+)} and Î›°_{m(Cl-)} are added and subtracted.

Î›°_{m (H2O)} = Î›°_{m(H+)} + Î›°_{m(OH-)}+ Î›°_{m(Na+)} - Î›°_{m(Na+)} + Î›°_{m(Cl-)} - Î›°_{m(Cl-)}

Î›°_{m (H2O)} = Î›°_{m(HCl)} + Î›°_{m(NaOH)} - Î›°_{m(NaCl)}

If we know the other values, Î›°_{m (H2O)} can be calculated.

**Question 3.**The molar conductivity of 0.025 mol L^{–1} methanoic acid is 46.1 S cm^{2} mol^{–1}. Calculate its degree of dissociation and dissociation constant. Given Î»^{0}_{(H+)} = 349.6 S cm^{2} mol^{–1} and Î»^{0} (HCOO^{–}) = 54.6 S cm^{2} mol^{–1}.

**Answer:**C = 0.025 mol L^{-1}

A_{m} = 46.1 Scm^{2} mol^{-1}

Î»^{0} (H^{+}) = 349.6 Scm^{2} mol^{-1}

Î»^{0} (HCOO^{-}) = 54.6 Scm^{2} mol^{-1}

Î›^{0}_{m} (HCOOH) = Î› ^{0}(H^{+}) + Î›^{0}(HCOO^{-})

= 349.6 + 54.6

= 404.2 S cm^{2} mol^{-1}

Now, the degree of dissociation:

∴ Î± = 0.114(approximately)

Thus, dissociation constant:

⇒ K = 3.67 × 10^{-4} mol L^{-1}

**Question 1.**

Why does the conductivity of a solution decrease with dilution?

**Answer:**

The conductivity of a solution depends on the amount of ions present per volume of the solution. When diluted, the concentration of the ions decreases which implies that the number of ions per volume decreases thus, in turn, conductivity decreases.

**Question 2.**

Suggest a way to determine the Î›°_{m} value of water

**Answer:**

Î›°_{m (H2O)} = Î›°_{m(H+)} + Î›°_{m(OH-)}

For calculation purpose, Î›°_{m(Na+)} and Î›°_{m(Cl-)} are added and subtracted.

Î›°_{m (H2O)} = Î›°_{m(H+)} + Î›°_{m(OH-)}+ Î›°_{m(Na+)} - Î›°_{m(Na+)} + Î›°_{m(Cl-)} - Î›°_{m(Cl-)}

Î›°_{m (H2O)} = Î›°_{m(HCl)} + Î›°_{m(NaOH)} - Î›°_{m(NaCl)}

If we know the other values, Î›°_{m (H2O)} can be calculated.

**Question 3.**

The molar conductivity of 0.025 mol L^{–1} methanoic acid is 46.1 S cm^{2} mol^{–1}. Calculate its degree of dissociation and dissociation constant. Given Î»^{0}_{(H+)} = 349.6 S cm^{2} mol^{–1} and Î»^{0} (HCOO^{–}) = 54.6 S cm^{2} mol^{–1}.

**Answer:**

C = 0.025 mol L^{-1}

A_{m} = 46.1 Scm^{2} mol^{-1}

Î»^{0} (H^{+}) = 349.6 Scm^{2} mol^{-1}

Î»^{0} (HCOO^{-}) = 54.6 Scm^{2} mol^{-1}

Î›^{0}_{m} (HCOOH) = Î› ^{0}(H^{+}) + Î›^{0}(HCOO^{-})

= 349.6 + 54.6

= 404.2 S cm^{2} mol^{-1}

Now, the degree of dissociation:

∴ Î± = 0.114(approximately)

Thus, dissociation constant:

⇒ K = 3.67 × 10^{-4} mol L^{-1}

###### Intext Questions Pg-86

**Question 1.**If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

**Answer:**Current I = 0.5A

Time t = 2hrs = 2×60×60 = 7200 seconds

Charge __Q = I×t__

Q = 0.5×7200 = 3600 C

Charge carried by 1 mole of electrons (6.023×10^{23}electrons) is equal to 96487C.

No of electrons = 6.023×10^{23} × 3600/96487

No of electrons = 2.25×10^{22} electrons.

**Question 2.**Suggest a list of metals that are extracted electrolytically.

**Answer:**Metals with greater reactivity can be extracted electrolytically. Sodium, potassium, calcium, lithium, magnesium, aluminium which are present in the top of the reactivity series are extracted electrolytically.

**Question 3.**Consider the reaction: Cr_{2}O_{7}^{2–} + 14H^{+} + 6e^{–}⇒ 2Cr^{3+} + 7H_{2}O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_{2}O_{7}^{2–}?

**Answer:**Cr_{2}O_{7}^{2–} + 14H^{+} + 6e^{–}⇒ 2Cr^{3+} + 7H_{2}O

For reducing one mole of Cr_{2}O_{7}^{2–}, 6 mole of electrons are required. Hence, 6 faraday charge is needed. Hence, 6F = 6×96487 = 578922 C. Thus, the quantity of electricity is needed is 578922 C.

**Question 1.**

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

**Answer:**

Current I = 0.5A

Time t = 2hrs = 2×60×60 = 7200 seconds

Charge __Q = I×t__

Q = 0.5×7200 = 3600 C

Charge carried by 1 mole of electrons (6.023×10^{23}electrons) is equal to 96487C.

No of electrons = 6.023×10^{23} × 3600/96487

No of electrons = 2.25×10^{22} electrons.

**Question 2.**

Suggest a list of metals that are extracted electrolytically.

**Answer:**

Metals with greater reactivity can be extracted electrolytically. Sodium, potassium, calcium, lithium, magnesium, aluminium which are present in the top of the reactivity series are extracted electrolytically.

**Question 3.**

Consider the reaction: Cr_{2}O_{7}^{2–} + 14H^{+} + 6e^{–}⇒ 2Cr^{3+} + 7H_{2}O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_{2}O_{7}^{2–}?

**Answer:**

Cr_{2}O_{7}^{2–} + 14H^{+} + 6e^{–}⇒ 2Cr^{3+} + 7H_{2}O

For reducing one mole of Cr_{2}O_{7}^{2–}, 6 mole of electrons are required. Hence, 6 faraday charge is needed. Hence, 6F = 6×96487 = 578922 C. Thus, the quantity of electricity is needed is 578922 C.

###### Intext Questions Pg-90

**Question 1.**Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging

**Answer:**Anode: Lead (Pb)

Cathode: a grid of lead packed with lead oxide (PbO_{2})

Electrolyte: 38% solution of sulphuric acid (H_{2}SO_{4})

The cell reactions are as follows :

Pb (s) + SO_{2}^{-4}(aq) ⇒ PbSO_{4}(s) + 2e^{-} (anode)

PbO_{2}(s) + SO_{2}^{-4}(aq) + 4H^{+}(aq) +2e^{-}⇒ PbSO_{4}(s) +2H_{2}O(l) (cathode)

Pb(s) + PbO_{2}(s) +2H_{2}SO_{4}(aq)⇒ 2PbSO_{4}(s) +2H_{2}O(l)

(overall cell reaction)

On charging, all these reactions will be reversed.

**Question 2.**Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

**Answer:**Methane, oxygen & methanol can be used as fuels in fuel cells other than hydrogen.

**Question 3.**Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

**Answer:**In the corrosion reaction, due to the presence of air and moisture, oxidation takes place at a particular point of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe(s) ⇒ Fe^{2+}_{(aq)} + 2e^{-}

Electrons released at the anodic spot move through the metal and go to another spot of the object, wherein presence of H^{+} ions, the electrons reduce oxygen. This spot behaves as the cathode. These H^{+} ions come either from H_{2}CO_{3}, which are formed due to the dissolution of carbon dioxide from the air into water. The cathodic reaction is given by

O_{2(air)} + 4H_{(aq)}^{+}+4e^{-}⇒ 2H_{2}O

The overall reaction is given by,

2Fe_{(s)} + O_{2(air)} + 4H_{(aq)}^{+}⇒ 2Fe^{2+}+2H_{2}O

**Question 1.**

Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging

**Answer:**

Anode: Lead (Pb)

Cathode: a grid of lead packed with lead oxide (PbO_{2})

Electrolyte: 38% solution of sulphuric acid (H_{2}SO_{4})

The cell reactions are as follows :

Pb (s) + SO_{2}^{-4}(aq) ⇒ PbSO_{4}(s) + 2e^{-} (anode)

PbO_{2}(s) + SO_{2}^{-4}(aq) + 4H^{+}(aq) +2e^{-}⇒ PbSO_{4}(s) +2H_{2}O(l) (cathode)

Pb(s) + PbO_{2}(s) +2H_{2}SO_{4}(aq)⇒ 2PbSO_{4}(s) +2H_{2}O(l)

(overall cell reaction)

On charging, all these reactions will be reversed.

**Question 2.**

Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

**Answer:**

Methane, oxygen & methanol can be used as fuels in fuel cells other than hydrogen.

**Question 3.**

Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

**Answer:**

In the corrosion reaction, due to the presence of air and moisture, oxidation takes place at a particular point of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe(s) ⇒ Fe^{2+}_{(aq)} + 2e^{-}

Electrons released at the anodic spot move through the metal and go to another spot of the object, wherein presence of H^{+} ions, the electrons reduce oxygen. This spot behaves as the cathode. These H^{+} ions come either from H_{2}CO_{3}, which are formed due to the dissolution of carbon dioxide from the air into water. The cathodic reaction is given by

O_{2(air)} + 4H_{(aq)}^{+}+4e^{-}⇒ 2H_{2}O

The overall reaction is given by,

2Fe_{(s)} + O_{2(air)} + 4H_{(aq)}^{+}⇒ 2Fe^{2+}+2H_{2}O

###### Exercises

**Question 1.**Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg, and Zn.

**Answer:**The order in which the given metals displace each other from the solution of their salts is given by,

Mg>Al> Zn> Fe> Cu

Explanation -

A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt. The order of increasing the reducing power of given metals is Cu< Fe< Zn< Al<Mg. Hence, we can interpret as Mg can displace Cu from its salt solution, but Cu cannot displace Mg. The order in which the given metals displace each other from the solution of their salts is given by, Mg>Al> Zn> Fe> Cu. This is hence arranged in decreasing order of its reactivity.

**Question 2.**Given the standard electrode potentials,

K^{+} /K = –2.93V, Ag^{+} /Ag = 0.80V,

Hg^{2+} /Hg = 0.79V

Mg^{2+} /Mg = –2.37 V, Cr^{3+} /Cr = – 0.74V

Arrange these metals in their increasing order of reducing power.

**Answer:**Reducing power of metals increase with the decrease of reduction potential. Hence, the increasing order of reducing power will be as,

Ag < Hg < Cr < Mg < K

Explanation -

When the reduction potential is lower, the element has more tendency to get oxidized and thus more will be reducing power. The metal that has more negative electrode potential will be the one with more reducing power. Thus, here potassium(K) has the highest reducing power among the given elements.

**Question 3.**Depict the galvanic cell in which the reaction

Zn(s) + 2Ag^{+} (aq) → Zn^{2+} (aq) + 2Ag(s) takes place. Further show:

1) Which of the electrode is negatively charged?

2) The carriers of the current in the cell.

3) Individual reaction at each electrode.

**Answer:**The galvanic cell corresponding to the given redox reaction can be represented as:

Zn|Zn^{2 +}_{(aq)}||Ag ^{+}_{(aq)}|Ag

1) __Zn electrode (anode) is negatively charged__ because, at this electrode, Zn is oxidized to Zn^{2+}, causing electron accumulation at the anode.

2) __Electrons (ions) are the carriers__ of the current in the cell and in the external circuit, current flows from Ag (cathode) to Zn(anode) which is normally opposite to the electron flow which is from anode to cathode.

3) At anode:

Zn_{(s)}⇒ Zn^{2 +}_{(aq)} + 2e^{–}

At cathode:

Ag ^{+}_{(aq)} + e ^{–}⇒ Ag_{(s)}

**Question 4.**Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(i) 2Cr(s) + 3Cd^{2+} (aq) → 2Cr^{3+} (aq) + 3Cd

(ii) Fe^{2} + (aq) + Ag^{+} (aq) → Fe^{3+} (aq) + Ag(s)

Calculate the Î”_{r}G^{0} and equilibrium constant(K) of the reactions.

**Answer:**(i) Known - E^{0}_{Cr}^{3 +}_{/Cr} = - 0.74 V

E^{0}_{Cd}^{2 +}_{/Cd} = - 0.40 V

∆_{r}G^{0} = ?

K = ?

The galvanic cell of the given reaction is written as -

Cr_{(s)}|Cr^{3 +}_{(aq)}|| Cd^{2 +}_{(aq)}|Cd_{(s)}→ Reaction 1

Hence, the standard cell potential is given as,

E^{0} = E^{0}_{R} - E^{0}_{L}

= - 0.40 - (- 0.74)

∴ E^{0} = + 0.34 V

To calculate the standard Gibb’s free energy, ∆_{r}G^{0}, we use,

∆_{r}G^{0} = - nE^{0}F → Equation 1

where nF is the amount of charge passed and E^{0} is the standard reduction electrode potential.

Substituting n = 6 (no. of e ^{-} involved in the reaction 1), F = 96487 C mol^{-1},

E^{0} = + 0.34 V in Equation 1, we get,

∆_{r}G^{0} = - 6×0.34V×96487 C mol^{-1}

= - 196833.48 CV mol^{-1}

= - 196833.48 J mol^{-1}

∴ ∆_{r}G^{0} = - 196.83348 kJ mol^{-1}

To find out the equilibrium constant, K, we use the formula,

log K = 34.5177

K = antilog 34.5177

∴ K = 3.294 × 10^{34}

__The____standard Gibb’s free energy, ∆___{r}G^{0} is - 196.83348 kJ mol ^{– 1} and equilibrium constant, K is__3.294 × 10__^{34}

(ii) Known -

E^{0}_{Fe}^{3 +}_{/Fe}^{2 +} = 0.77V

E^{0}_{Ag}^{+}_{/Ag} = 0.80 V

∆_{r}G^{0} = ?

K = ?

The galvanic cell of the given reaction is written as -

Fe^{2 +}_{(aq)}|Fe^{3 +}_{(aq)}|| Ag ^{+}_{(aq)}|Ag_{(s)}→ Reaction 1

Hence, the standard cell potential is given as,

E^{0} = E^{0}_{R} - E^{0}_{L}

= 0.80 - (0.77)

∴ E^{0} = + 0.03 V

To calculate the standard Gibb’s free energy, ∆_{r}G^{0}, we use,

∆_{r}G^{0} = - nE^{0}F → Equation 1

where nF is the amount of charge passed and E^{0} is the standard reduction electrode potential.

Substituting n = 1 (no. of e ^{-} involved in the reaction 1), F = 96487 C mol^{-1}, E^{0} = + 0.03V in Equation 1, we get,

∆_{r}G^{0} = - 1×0.03V×96487 C mol^{-1}

= - 2894.61 CV mol ^{- 1}

= - 2894.61 J mol^{-1}

∴ ∆_{r}G^{0} = - 2.89461 kJ mol^{-1}

To find out the equilibrium constant, K, we use the formula,

log K = 0.5076

K = Antilog 0.5076

∴ K = 3.218

__The____standard Gibb’s free energy, ∆___{r}G^{0} is - 2.89461 kJ mol ^{– 1} and equilibrium constant, K is__3.218__

**Question 5.**Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s)|Mg^{2+} (0.001M)||Cu^{2+} (0.0001 M)|Cu(s)

(ii) Fe(s)|Fe^{2+} (0.001M)||H^{+} (1M)|H_{2}(g)(1bar)| Pt(s)

(iii) Sn(s)|Sn^{2+} (0.050 M)||H^{+} (0.020 M)|H_{2}(g) (1 bar)|Pt(s)

(iv) Pt(s)|Br_{2}(l)|Br^{–}(0.010 M)||H^{+} (0.030 M)| H_{2}(g) (1 bar)|Pt(s).

**Answer:**E_{cell} = ?

(i) Mg + Cu^{2 +} → Mg^{2 +} + Cu (n = 2)

E^{0}_{Cu}^{2 +}_{/Cu}^{+} = 0.34V

E^{0}_{Mg}^{2 +}_{/Mg} = - 2.37 V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0.34 - ( - 2.37) → Equation 1

Using Nernst equation, we get,

→ Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0.34 - ( - 2.37) -

= 2.71 -

= 2.71 - 0.02955

∴ E_{cell} = 2.68 V

__The e.m.f of the cell, E___{cell} is 2.68 V

ii) Fe + 2H ^{+} → Fe^{2 +} + H_{2} (n = 2)

E^{0}_{H}^{+}_{/H2} = 0 V

E^{0}_{Fe}^{2 +}_{/Fe} = - 0.44 V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - ( - 0.44) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - ( - 0.44) -

= 0.44 -

= 0.44 - 0.0887

∴ E_{cell} = 0.5287 V

__The e.m.f of the cell, E___{cell} is 0.5287 V

iii) Sn + 2H ^{+} → Sn^{2 +} + H_{2} (n = 2)

E^{0}_{H}^{+}_{/H2} = 0 V

E^{0}_{Sn}^{2 +}_{/Sn} = - 0.14V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - ( - 0.14) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - ( - 0.14) -

= 0.14 -

= 0.14 - 0.0295×2.0969

∴ E_{cell} = 0.08 V

__The e.m.f of the cell, E___{cell} is 0.08 V

iv) 2Br ^{-} + 2H ^{+} → Br_{2} + H_{2} (n = 2)

E^{0}_{H}^{+}_{/H2} = 0 V

E^{0}_{Br2/Br}^{-} = 1.08V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - (1.08) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - (1.08) -

= - 1.08 -

= - 1.08 - 0.208

∴ E_{cell} = 1.288 V

__The e.m.f of the cell, E___{cell} is 1.288 V

**Question 6.**In the button cells widely used in watches and other devices the following reaction takes place:

Zn_{(s)} + Ag_{2}O_{(s)} + H_{2}O_{(l)}→ Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{–}(aq). Determine Î”_{r}G^{0} and E^{0} for the reaction.

**Answer:**Given - Zn → Zn^{2 +} + 2e ^{-} , E^{0} = 0.76V (anode)

Ag_{2}O + H_{2}O + 2e ^{-} →2Ag + 2OH ^{-} , E^{0} = 0.344V (cathode), n = 2

Î”_{r}G^{0} = ?

E^{0}_{cell} = ?

Zn is oxidized and Ag_{2}O is reduced.

Hence, the standard cell potential, E^{0}_{cell} is given as,

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0.344 + 0.76

∴ E^{0}_{cell} = 1.104 V

To calculate the standard Gibb’s free energy, ∆_{r}G^{0}, we use,

∆_{r}G^{0} = - nE^{0}F → Equation 1

= - 2×96487×1.104 J

= - 213043.296 J

∴ ∆_{r}G^{0} = - 2.13×10^{5} J

__The standard cell potential,____E__^{0}_{cell} is 1.104 V and the__standard Gibb’s free energy, ∆___{r}G^{0} is__- 2.13×10__^{5} J

**Question 7.**Define conductivity and molar conductivity of the solution of an electrolyte. Discuss their variation with concentration.

**Answer:**__The conductivity____of a solution__ is defined as the conductance of one unit volume of solution kept between two platinum electrodes with a unit area of cross-section and at a distance of unit length.

__The molar conductivity____of the solution__ is defined as the conducting power of all the ions produced by one gram mole of an electrolyte in a solution. It is denoted by ∧_{m}.

__The conductivity____of a solution__ (both for strong and weak electrolytes) always __decreases__ with the __decrease in concentration of the electrolyte__ i.e., on dilution. This pattern is seen because the number of ions per unit volume that carry the current in the solution decreases on dilution. __The molar conductivity of the solution increases__ with the __decrease in concentration of the electrolyte__. This is because both the number of ions as well as mobility of ions increase with dilution.

**Question 8.**The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^{–1}. Calculate its molar conductivity.

**Answer:**Given -

Molarity, C = 0.20 M

Electrolytic conductivity of a solution, Îº = 0.0248 S cm^{–1}

Molar conductivity = ?

Molar conductivity, ∧_{m} = S cm^{2} mol^{-1}

∴ ∧_{m} = 124 S cm^{2} mol^{-1}

__Molar conductivity(____∧___{m}__) of 0.20 M solution of KCl at 298 K is 124 S____cm__^{2} mol^{-1}

**Question 9.**The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Î©. What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^{–3} S cm^{–1}.

**Answer:**Given -

Resistance of a conductivity cell, R = 1500

Electrolytic conductivity of a solution, Îº = 0.146 × 10^{–3} S cm^{–1}

Cell constant = ?

Conductivity, Îº =

Cell constant = Îº × R

= 0.146 × 10^{–3} S cm^{–1}×1500

Cell constant = 0.219 cm^{-1}

__The cell constant of the cell containing 0.001M KCl solution at 298 K is 0.219 cm__^{-1}

**Question 10.**The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Calculate Î›_{m} for all concentrations and draw a plot between Î›_{m} and c^{1/2}. Find the value of Î›^{0}_{m.}

**Answer:**(unit conversion factor)

__∧__^{0}_{m}__= Intercept on the____∧___{m}__axis = 124.0 S cm__^{2} mol^{-1}, which is obtained by extrapolation to zero concentration.

**Question 11.**Conductivity of 0.00241 M acetic acid is 7.896 × 10^{–5} S cm^{–1}. Calculate its molar conductivity. If ∧^{0}_{m} for acetic acid is 390.5 S cm^{2}mol^{–1}, what is its dissociation constant?

**Answer:**Given -

Molarity, C = 0.00241 M

Conductivity, Îº = 7.896 × 10^{–5} S cm^{–1}

Molar conductivity, ∧_{m} = ?

for acetic acid = 390.5 S cm^{2}mol^{–1}

Molar conductivity, ∧_{m} = S cm^{2} mol^{-1}

∴ ∧_{m} = 32.76 S cm^{2} mol^{-1}

To calculate the dissociation constant, K_{a}, we use

K_{a} = → Equation 1

Here, we need to find the value of Î± (degree of dissociation), by the formula,

∴ Î± = 8.4×10^{-2} ⇒ Equation 2

→ Thus, substituting Equation 2 in Equation 1, we get,

K_{a} =

=

∴ K_{a} = 1.86×10 ^{- 5}

__The____molar conductivity,____∧___{m}__is____32.76____S cm__^{2} mol^{-1} and the__dissociation constant, K___{a} is__1.86×10__^{-5}

**Question 12.**How much charge is required for the following reductions:

(i) 1 mol of Al^{3 +} to Al?

(ii) 1 mol of Cu^{2 +} to Cu?

(iii) 1 mol of MnO^{–}_{4} to Mn^{2+} ?

**Answer:**(i) The electrode reaction is given as,

Al^{3 +}_{(aq)} + 3e ^{-} → Al_{(s)}

∴ The quantity of charge required for the reduction of 1 mol of Al^{3+} = 3F

= 3×96487 C

= 289461 C

(ii) The electrode reaction is given as,

Cu^{2 +}_{(aq)} + 2e ^{-} → Cu_{(s)}

∴ The quantity of charge required for the reduction of 1 mol of Cu^{2+} = 2F

= 2×96487 C

= 192974 C

(iii) The electrode reaction is given as,

MnO_{4}→ Mn^{2 +}

i.e., Mn^{7 +} + 5e ^{-} → Mn^{2 +}

∴ The quantity of charge required for the reduction of 1 mol of Mn^{7+} = 5F

= 5×96487 C

= 482435 C

**Question 13.**How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl_{2}?

(ii) 40.0 g of Al from molten Al_{2}O_{3}?

**Answer:**(i) Ca^{2+} + 2e^{-} → Ca

⇒ Here, 1 mole of Ca, i.e., 40g of Ca requires = 2 F electricity (F if Faraday)

∴ 20g of Ca requires =

= 1 F of electricity

__E____lectricity in terms of Faraday required to produce 20.0 g of Ca from molten CaCl___{2} is__1 F of electricity.__

(ii) Al^{3 +} + 3e ^{-} → Al

⇒ 1 mole of Al, i.e., 27g of Al requires = 3 F electricity (F if Faraday)

∴ 40.0 g of Al will require =

= 4.44 F of electricity

__E____lectricity in terms of Faraday required to produce 40.0 g of Al from molten Al___{2}O_{3} is 4.__44 F of electricity__

**Question 14.**How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H_{2}O to O_{2}?

(ii) 1 mol of FeO to Fe_{2}O_{3}?

**Answer:**(i)The electrode reaction for 1 mole of H_{2}O is given as,

H_{2}O → H_{2} + O_{2}

i.e., O^{2 -} →O_{2} + 2e ^{-}

∴ The quantity of electricity required = 2F

= 2×96487 C

= 192974 C

__T____he quantity of electricity required in coulomb for the oxidation of 1 mol of H___{2}O to O_{2} is 192974 C

(ii) The electrode reaction for 1 mole of FeO is

FeO + O_{2} → Fe_{2}O_{3}

i.e., Fe^{2 +} → Fe^{3 +} + e ^{-}

∴ The quantity of electricity required = 1F

= 1×96487 C

= 96487 C

__T____he quantity of electricity required in coulomb for the oxidation of 1 mol of FeO to Fe___{2}O_{3} is 96487 C

**Question 15.**A solution of Ni(NO_{3})_{2} is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What is mass of Ni deposited at the cathode?

Given -

Current = 5A

Time - 20 minutes

Mass of Ni deposited = ?

**Answer:**Quantity of electricity passed = 5 A × (20 × 60 sec)

= 6000 C ⇒ Equation 1

The electrode reaction is written as,

Ni^{2 +} + 2e → Ni

Thus, the quantity of electricity required = 2F

= 2×96487 C

= 192974 C

∵ 192974 C of electricity deposits 1 mole of Ni, which is 58.7 g ⇒ Equation 2

Thus, equating equations 1 and 2, we get,

192974 C of electricity deposits = 58.7 g

6000 C of electricity will deposit =

= 1.825g of Ni

__The mass of Ni deposited at the cathode is 1.825g of Ni__

**Question 16.**Three electrolytic cells A,B,C containing solutions of ZnSO_{4}, AgNO_{3} and CuSO_{4}, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Given -

I = 1.5 A

W = 1.45 g of Ag

t = ?

n = 1

**Answer:**Equivalent weight is Ag, E_{Ag} = = 180

Equivalent weight is Cu, E_{Cu} = = 31.75

Equivalent weight is Zn, E_{Zn}= = 32.5

Using Faraday’s second law of electrolysis, to find the mass of Cu and Zn, we use Equation 1,

→ Equation 1

⇒

∴ W_{Cu} = 0.426 g

⇒

∴ W_{Zn} = 0.436 g

To find the time of current flow, using Faraday’s first law of electrolysis we get,

M = Z ×I ×t ⇒ Equation 2

∵ Z = , Equation 2 becomes,

M =

t =

t = 864 seconds.

__The time of current flow, t = 864 seconds, the mass of Cu is 0.426 g and mass of Zn is 0.436 g__

**Question 17.**Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe^{3 +} (aq) and I^{–}(aq)

(ii) Ag ^{+} (aq) and Cu(s)

(iii) Fe^{3 +} (aq) and Br^{–} (aq)

(iv) Ag(s) and Fe^{3 +} (aq)

(v) Br_{2} (aq) and Fe^{2 +} (aq).

**Answer:**(i) The electrode reaction is written as,

2Fe^{3 +} + 2I ^{-} → 2Fe^{2 +} + I_{2}

E^{0}_{cell} =

= 0.54V - 0.77V

∴ E^{0}_{cell} = - 0.23 V

It is not feasible, as E^{0}_{cell} is negative, ∴ ∆G^{0} is positive.

(ii) The electrode reaction is written as,

2Ag ^{+}_{(aq)} + Cu_{(s)}→ Cu^{2 +}_{(aq)} + Ag_{(s)}

E^{0}_{cell} =

= + 0.80V - 0.34V

∴ E^{0}_{cell} = 0.46V

It is feasible, as E^{0}_{cell} is positive, ∴ ∆G^{0} is negative.

(iii) The electrode reaction is written as,

2Fe^{3 +}_{(aq)} + 2Br ^{-}_{(aq)}→ 2Fe^{2 +}_{(aq)} + Br_{2}

E^{0}_{cell} =

= 0.77V - 1.09V

∴ E^{0}_{cell} = - 0.32 V

It is not feasible, as E^{0}_{cell} is negative, ∴ ∆G^{0} is positive.

(iv) The electrode reaction is written as,

Ag_{(s)} + Fe^{3 +}_{(aq)} → Fe^{2 +}_{(aq)} + Ag ^{+}_{(aq)}

E^{0}_{cell} =

= 0.77V - 0.80V

∴ E^{0}_{cell} = - 0.03

It is not feasible, as E^{0}_{cell} is negative, ∴ ∆G^{0} is positive.

(v) The electrode reaction is written as,

Br_{2} + 2Fe^{2 +}_{(aq)} → 2Br ^{-}_{(aq)} + 2Fe^{3 +}_{(aq)}

E^{0}_{cell} =

= 1.09V - 0.77V

∴ E^{0}_{cell} = 0.32 V

It is feasible, as E^{0}_{cell} is positive, ∴ ∆G^{0} is negative.

**Question 18.**Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO_{3} with silver electrodes.

(ii) An aqueous solution of AgNO_{3} with platinum electrodes.

(iii) A dilute solution of H_{2}SO_{4} with platinum electrodes.

(iv) An aqueous solution of CuCl_{2} with platinum electrodes.

**Answer:**Given -

All the ions are in aqueous state.

(i) Reaction in solution:

AgNO_{3(s)} + aq → Ag ^{+} + NO^{3 -}

H_{2}O Ã³ H ^{+} + OH ^{-}

At cathode:

Ag ^{+}_{(aq)} + e ^{-} →Ag_{(s)}

Ag ^{+} ions have lower discharge potential than H ^{+} ions. Hence, Ag ^{+} ions get deposited as Ag in preference to H ^{+} ions.

At anode:

Ag_{(s)}→ Ag ^{+}_{(aq)} + e ^{-}

As Ag anode is attacked by NO^{3 -} ions, Ag of the anode will dissolve to form Ag ^{+} ions in the aqueous solution.

(ii) Reaction in solution:

AgNO_{3(s)} + aq → Ag ^{+} + NO^{3 -}

H_{2}O Ã³H ^{+} + OH ^{-}

At cathode:

2Ag ^{+}_{(aq)} + 2e ^{-} →2Ag_{(s)}

Ag ^{+} ions have lower discharge potential than H ^{+} ions. Hence, Ag ^{+} ions get deposited as Ag in preference to H ^{+} ions.

At anode:

2OH ^{-}_{(aq)} → O_{2(g)} + 2H ^{+}_{(aq)} + 4e ^{-}

As anode is not attackable, out of OH ^{-} and NO^{3 -} ions, OH ^{-} having lower discharge potential, will be discharged in preference to NO^{3 -} ions. These then decompose to give out O_{2}.

(iii) Reaction in solution:

H_{2}SO_{4(aq)} → 2H ^{+}_{(aq)} + SO_{4}^{2 -}_{(aq)}

At cathode:

2H ^{+}_{(aq)} + 2e ^{-} →H_{2(g)}

At anode:

2OH ^{-}_{(aq)} → O_{2(g)} + 2H ^{+}_{(aq)} + 4e ^{-}

__∴____H___{2} gas is evolved at cathode and O_{2(g)} is evolved at anode.

(iv) Reaction in solution:

CuCl_{2(s)} + aq → Cu^{2 +}_{(aq)} + Cl ^{-}_{(aq)}

H_{2}O Ã³H ^{+} + OH ^{-}

At cathode:

Cu^{2 +}_{(aq)} + 2e ^{-} →Cu_{(g)}

At anode:

2Cl ^{-}_{(aq)} - 2e ^{-} → Cl_{2(g)}

__∴____Cu will be deposited at cathode and Cl___{2} gas will be liberated at anode.

**Question 1.**

Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg, and Zn.

**Answer:**

The order in which the given metals displace each other from the solution of their salts is given by,

Mg>Al> Zn> Fe> Cu

Explanation -

A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt. The order of increasing the reducing power of given metals is Cu< Fe< Zn< Al<Mg. Hence, we can interpret as Mg can displace Cu from its salt solution, but Cu cannot displace Mg. The order in which the given metals displace each other from the solution of their salts is given by, Mg>Al> Zn> Fe> Cu. This is hence arranged in decreasing order of its reactivity.

**Question 2.**

Given the standard electrode potentials,

K^{+} /K = –2.93V, Ag^{+} /Ag = 0.80V,

Hg^{2+} /Hg = 0.79V

Mg^{2+} /Mg = –2.37 V, Cr^{3+} /Cr = – 0.74V

Arrange these metals in their increasing order of reducing power.

**Answer:**

Reducing power of metals increase with the decrease of reduction potential. Hence, the increasing order of reducing power will be as,

Ag < Hg < Cr < Mg < K

Explanation -

When the reduction potential is lower, the element has more tendency to get oxidized and thus more will be reducing power. The metal that has more negative electrode potential will be the one with more reducing power. Thus, here potassium(K) has the highest reducing power among the given elements.

**Question 3.**

Depict the galvanic cell in which the reaction

Zn(s) + 2Ag^{+} (aq) → Zn^{2+} (aq) + 2Ag(s) takes place. Further show:

1) Which of the electrode is negatively charged?

2) The carriers of the current in the cell.

3) Individual reaction at each electrode.

**Answer:**

The galvanic cell corresponding to the given redox reaction can be represented as:

Zn|Zn^{2 +}_{(aq)}||Ag ^{+}_{(aq)}|Ag

1) __Zn electrode (anode) is negatively charged__ because, at this electrode, Zn is oxidized to Zn^{2+}, causing electron accumulation at the anode.

2) __Electrons (ions) are the carriers__ of the current in the cell and in the external circuit, current flows from Ag (cathode) to Zn(anode) which is normally opposite to the electron flow which is from anode to cathode.

3) At anode:

Zn_{(s)}⇒ Zn^{2 +}_{(aq)} + 2e^{–}

At cathode:

Ag ^{+}_{(aq)} + e ^{–}⇒ Ag_{(s)}

**Question 4.**

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(i) 2Cr(s) + 3Cd^{2+} (aq) → 2Cr^{3+} (aq) + 3Cd

(ii) Fe^{2} + (aq) + Ag^{+} (aq) → Fe^{3+} (aq) + Ag(s)

Calculate the Î”_{r}G^{0} and equilibrium constant(K) of the reactions.

**Answer:**

(i) Known - E^{0}_{Cr}^{3 +}_{/Cr} = - 0.74 V

E^{0}_{Cd}^{2 +}_{/Cd} = - 0.40 V

∆_{r}G^{0} = ?

K = ?

The galvanic cell of the given reaction is written as -

Cr_{(s)}|Cr^{3 +}_{(aq)}|| Cd^{2 +}_{(aq)}|Cd_{(s)}→ Reaction 1

Hence, the standard cell potential is given as,

E^{0} = E^{0}_{R} - E^{0}_{L}

= - 0.40 - (- 0.74)

∴ E^{0} = + 0.34 V

To calculate the standard Gibb’s free energy, ∆_{r}G^{0}, we use,

∆_{r}G^{0} = - nE^{0}F → Equation 1

where nF is the amount of charge passed and E^{0} is the standard reduction electrode potential.

Substituting n = 6 (no. of e ^{-} involved in the reaction 1), F = 96487 C mol^{-1},

E^{0} = + 0.34 V in Equation 1, we get,

∆_{r}G^{0} = - 6×0.34V×96487 C mol^{-1}

= - 196833.48 CV mol^{-1}

= - 196833.48 J mol^{-1}

∴ ∆_{r}G^{0} = - 196.83348 kJ mol^{-1}

To find out the equilibrium constant, K, we use the formula,

log K = 34.5177

K = antilog 34.5177

∴ K = 3.294 × 10^{34}

__The____standard Gibb’s free energy, ∆ _{r}G^{0} is - 196.83348 kJ mol ^{– 1} and equilibrium constant, K is__

__3.294 × 10__

^{34}(ii) Known -

E^{0}_{Fe}^{3 +}_{/Fe}^{2 +} = 0.77V

E^{0}_{Ag}^{+}_{/Ag} = 0.80 V

∆_{r}G^{0} = ?

K = ?

The galvanic cell of the given reaction is written as -

Fe^{2 +}_{(aq)}|Fe^{3 +}_{(aq)}|| Ag ^{+}_{(aq)}|Ag_{(s)}→ Reaction 1

Hence, the standard cell potential is given as,

E^{0} = E^{0}_{R} - E^{0}_{L}

= 0.80 - (0.77)

∴ E^{0} = + 0.03 V

To calculate the standard Gibb’s free energy, ∆_{r}G^{0}, we use,

∆_{r}G^{0} = - nE^{0}F → Equation 1

where nF is the amount of charge passed and E^{0} is the standard reduction electrode potential.

Substituting n = 1 (no. of e ^{-} involved in the reaction 1), F = 96487 C mol^{-1}, E^{0} = + 0.03V in Equation 1, we get,

∆_{r}G^{0} = - 1×0.03V×96487 C mol^{-1}

= - 2894.61 CV mol ^{- 1}

= - 2894.61 J mol^{-1}

∴ ∆_{r}G^{0} = - 2.89461 kJ mol^{-1}

To find out the equilibrium constant, K, we use the formula,

log K = 0.5076

K = Antilog 0.5076

∴ K = 3.218

__The____standard Gibb’s free energy, ∆ _{r}G^{0} is - 2.89461 kJ mol ^{– 1} and equilibrium constant, K is__

__3.218__

**Question 5.**

Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s)|Mg^{2+} (0.001M)||Cu^{2+} (0.0001 M)|Cu(s)

(ii) Fe(s)|Fe^{2+} (0.001M)||H^{+} (1M)|H_{2}(g)(1bar)| Pt(s)

(iii) Sn(s)|Sn^{2+} (0.050 M)||H^{+} (0.020 M)|H_{2}(g) (1 bar)|Pt(s)

(iv) Pt(s)|Br_{2}(l)|Br^{–}(0.010 M)||H^{+} (0.030 M)| H_{2}(g) (1 bar)|Pt(s).

**Answer:**

E_{cell} = ?

(i) Mg + Cu^{2 +} → Mg^{2 +} + Cu (n = 2)

E^{0}_{Cu}^{2 +}_{/Cu}^{+} = 0.34V

E^{0}_{Mg}^{2 +}_{/Mg} = - 2.37 V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0.34 - ( - 2.37) → Equation 1

Using Nernst equation, we get,

→ Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0.34 - ( - 2.37) -

= 2.71 -

= 2.71 - 0.02955

∴ E_{cell} = 2.68 V

__The e.m.f of the cell, E _{cell} is 2.68 V__

ii) Fe + 2H ^{+} → Fe^{2 +} + H_{2} (n = 2)

E^{0}_{H}^{+}_{/H2} = 0 V

E^{0}_{Fe}^{2 +}_{/Fe} = - 0.44 V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - ( - 0.44) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - ( - 0.44) -

= 0.44 -

= 0.44 - 0.0887

∴ E_{cell} = 0.5287 V

__The e.m.f of the cell, E _{cell} is 0.5287 V__

iii) Sn + 2H ^{+} → Sn^{2 +} + H_{2} (n = 2)

E^{0}_{H}^{+}_{/H2} = 0 V

E^{0}_{Sn}^{2 +}_{/Sn} = - 0.14V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - ( - 0.14) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - ( - 0.14) -

= 0.14 -

= 0.14 - 0.0295×2.0969

∴ E_{cell} = 0.08 V

__The e.m.f of the cell, E _{cell} is 0.08 V__

iv) 2Br ^{-} + 2H ^{+} → Br_{2} + H_{2} (n = 2)

E^{0}_{H}^{+}_{/H2} = 0 V

E^{0}_{Br2/Br}^{-} = 1.08V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - (1.08) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - (1.08) -

= - 1.08 -

= - 1.08 - 0.208

∴ E_{cell} = 1.288 V

__The e.m.f of the cell, E _{cell} is 1.288 V__

**Question 6.**

In the button cells widely used in watches and other devices the following reaction takes place:

Zn_{(s)} + Ag_{2}O_{(s)} + H_{2}O_{(l)}→ Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{–}(aq). Determine Î”_{r}G^{0} and E^{0} for the reaction.

**Answer:**

Given - Zn → Zn^{2 +} + 2e ^{-} , E^{0} = 0.76V (anode)

Ag_{2}O + H_{2}O + 2e ^{-} →2Ag + 2OH ^{-} , E^{0} = 0.344V (cathode), n = 2

Î”_{r}G^{0} = ?

E^{0}_{cell} = ?

Zn is oxidized and Ag_{2}O is reduced.

Hence, the standard cell potential, E^{0}_{cell} is given as,

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0.344 + 0.76

∴ E^{0}_{cell} = 1.104 V

To calculate the standard Gibb’s free energy, ∆_{r}G^{0}, we use,

∆_{r}G^{0} = - nE^{0}F → Equation 1

= - 2×96487×1.104 J

= - 213043.296 J

∴ ∆_{r}G^{0} = - 2.13×10^{5} J

__The standard cell potential,____E ^{0}_{cell} is 1.104 V and the__

__standard Gibb’s free energy, ∆__

_{r}G^{0}is__- 2.13×10__

^{5}J**Question 7.**

Define conductivity and molar conductivity of the solution of an electrolyte. Discuss their variation with concentration.

**Answer:**

__The conductivity____of a solution__ is defined as the conductance of one unit volume of solution kept between two platinum electrodes with a unit area of cross-section and at a distance of unit length.

__The molar conductivity____of the solution__ is defined as the conducting power of all the ions produced by one gram mole of an electrolyte in a solution. It is denoted by ∧_{m}.

__The conductivity____of a solution__ (both for strong and weak electrolytes) always __decreases__ with the __decrease in concentration of the electrolyte__ i.e., on dilution. This pattern is seen because the number of ions per unit volume that carry the current in the solution decreases on dilution. __The molar conductivity of the solution increases__ with the __decrease in concentration of the electrolyte__. This is because both the number of ions as well as mobility of ions increase with dilution.

**Question 8.**

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^{–1}. Calculate its molar conductivity.

**Answer:**

Given -

Molarity, C = 0.20 M

Electrolytic conductivity of a solution, Îº = 0.0248 S cm^{–1}

Molar conductivity = ?

Molar conductivity, ∧_{m} = S cm^{2} mol^{-1}

∴ ∧_{m} = 124 S cm^{2} mol^{-1}

__Molar conductivity(____∧___{m}__) of 0.20 M solution of KCl at 298 K is 124 S____cm ^{2} mol^{-1}__

**Question 9.**

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Î©. What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^{–3} S cm^{–1}.

**Answer:**

Given -

Resistance of a conductivity cell, R = 1500

Electrolytic conductivity of a solution, Îº = 0.146 × 10^{–3} S cm^{–1}

Cell constant = ?

Conductivity, Îº =

Cell constant = Îº × R

= 0.146 × 10^{–3} S cm^{–1}×1500

Cell constant = 0.219 cm^{-1}

__The cell constant of the cell containing 0.001M KCl solution at 298 K is 0.219 cm ^{-1}__

**Question 10.**

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Calculate Î›_{m} for all concentrations and draw a plot between Î›_{m} and c^{1/2}. Find the value of Î›^{0}_{m.}

**Answer:**

(unit conversion factor)

__∧__^{0}_{m}__= Intercept on the____∧___{m}__axis = 124.0 S cm ^{2} mol^{-1}, which is obtained by extrapolation to zero concentration.__

**Question 11.**

Conductivity of 0.00241 M acetic acid is 7.896 × 10^{–5} S cm^{–1}. Calculate its molar conductivity. If ∧^{0}_{m} for acetic acid is 390.5 S cm^{2}mol^{–1}, what is its dissociation constant?

**Answer:**

Given -

Molarity, C = 0.00241 M

Conductivity, Îº = 7.896 × 10^{–5} S cm^{–1}

Molar conductivity, ∧_{m} = ?

for acetic acid = 390.5 S cm^{2}mol^{–1}

Molar conductivity, ∧_{m} = S cm^{2} mol^{-1}

∴ ∧_{m} = 32.76 S cm^{2} mol^{-1}

To calculate the dissociation constant, K_{a}, we use

K_{a} = → Equation 1

Here, we need to find the value of Î± (degree of dissociation), by the formula,

∴ Î± = 8.4×10^{-2} ⇒ Equation 2

→ Thus, substituting Equation 2 in Equation 1, we get,

K_{a} =

=

∴ K_{a} = 1.86×10 ^{- 5}

__The____molar conductivity,____∧___{m}__is____32.76____S cm ^{2} mol^{-1} and the__

__dissociation constant, K__

_{a}is__1.86×10__

^{-5}**Question 12.**

How much charge is required for the following reductions:

(i) 1 mol of Al^{3 +} to Al?

(ii) 1 mol of Cu^{2 +} to Cu?

(iii) 1 mol of MnO^{–}_{4} to Mn^{2+} ?

**Answer:**

(i) The electrode reaction is given as,

Al^{3 +}_{(aq)} + 3e ^{-} → Al_{(s)}

∴ The quantity of charge required for the reduction of 1 mol of Al^{3+} = 3F

= 3×96487 C

= 289461 C

(ii) The electrode reaction is given as,

Cu^{2 +}_{(aq)} + 2e ^{-} → Cu_{(s)}

∴ The quantity of charge required for the reduction of 1 mol of Cu^{2+} = 2F

= 2×96487 C

= 192974 C

(iii) The electrode reaction is given as,

MnO_{4}→ Mn^{2 +}

i.e., Mn^{7 +} + 5e ^{-} → Mn^{2 +}

∴ The quantity of charge required for the reduction of 1 mol of Mn^{7+} = 5F

= 5×96487 C

= 482435 C

**Question 13.**

How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl_{2}?

(ii) 40.0 g of Al from molten Al_{2}O_{3}?

**Answer:**

(i) Ca^{2+} + 2e^{-} → Ca

⇒ Here, 1 mole of Ca, i.e., 40g of Ca requires = 2 F electricity (F if Faraday)

∴ 20g of Ca requires =

= 1 F of electricity

__E____lectricity in terms of Faraday required to produce 20.0 g of Ca from molten CaCl _{2} is__

__1 F of electricity.__

(ii) Al^{3 +} + 3e ^{-} → Al

⇒ 1 mole of Al, i.e., 27g of Al requires = 3 F electricity (F if Faraday)

∴ 40.0 g of Al will require =

= 4.44 F of electricity

__E____lectricity in terms of Faraday required to produce 40.0 g of Al from molten Al _{2}O_{3} is 4.__

__44 F of electricity__

**Question 14.**

How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H_{2}O to O_{2}?

(ii) 1 mol of FeO to Fe_{2}O_{3}?

**Answer:**

(i)The electrode reaction for 1 mole of H_{2}O is given as,

H_{2}O → H_{2} + O_{2}

i.e., O^{2 -} →O_{2} + 2e ^{-}

∴ The quantity of electricity required = 2F

= 2×96487 C

= 192974 C

__T____he quantity of electricity required in coulomb for the oxidation of 1 mol of H _{2}O to O_{2} is 192974 C__

(ii) The electrode reaction for 1 mole of FeO is

FeO + O_{2} → Fe_{2}O_{3}

i.e., Fe^{2 +} → Fe^{3 +} + e ^{-}

∴ The quantity of electricity required = 1F

= 1×96487 C

= 96487 C

__T____he quantity of electricity required in coulomb for the oxidation of 1 mol of FeO to Fe _{2}O_{3} is 96487 C__

**Question 15.**

A solution of Ni(NO_{3})_{2} is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What is mass of Ni deposited at the cathode?

Given -

Current = 5A

Time - 20 minutes

Mass of Ni deposited = ?

**Answer:**

Quantity of electricity passed = 5 A × (20 × 60 sec)

= 6000 C ⇒ Equation 1

The electrode reaction is written as,

Ni^{2 +} + 2e → Ni

Thus, the quantity of electricity required = 2F

= 2×96487 C

= 192974 C

∵ 192974 C of electricity deposits 1 mole of Ni, which is 58.7 g ⇒ Equation 2

Thus, equating equations 1 and 2, we get,

192974 C of electricity deposits = 58.7 g

6000 C of electricity will deposit =

= 1.825g of Ni

__The mass of Ni deposited at the cathode is 1.825g of Ni__

**Question 16.**

Three electrolytic cells A,B,C containing solutions of ZnSO_{4}, AgNO_{3} and CuSO_{4}, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Given -

I = 1.5 A

W = 1.45 g of Ag

t = ?

n = 1

**Answer:**

Equivalent weight is Ag, E_{Ag} = = 180

Equivalent weight is Cu, E_{Cu} = = 31.75

Equivalent weight is Zn, E_{Zn}= = 32.5

Using Faraday’s second law of electrolysis, to find the mass of Cu and Zn, we use Equation 1,

→ Equation 1

⇒

∴ W_{Cu} = 0.426 g

⇒

∴ W_{Zn} = 0.436 g

To find the time of current flow, using Faraday’s first law of electrolysis we get,

M = Z ×I ×t ⇒ Equation 2

∵ Z = , Equation 2 becomes,

M =

t =

t = 864 seconds.

__The time of current flow, t = 864 seconds, the mass of Cu is 0.426 g and mass of Zn is 0.436 g__

**Question 17.**

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe^{3 +} (aq) and I^{–}(aq)

(ii) Ag ^{+} (aq) and Cu(s)

(iii) Fe^{3 +} (aq) and Br^{–} (aq)

(iv) Ag(s) and Fe^{3 +} (aq)

(v) Br_{2} (aq) and Fe^{2 +} (aq).

**Answer:**

(i) The electrode reaction is written as,

2Fe^{3 +} + 2I ^{-} → 2Fe^{2 +} + I_{2}

E^{0}_{cell} =

= 0.54V - 0.77V

∴ E^{0}_{cell} = - 0.23 V

It is not feasible, as E^{0}_{cell} is negative, ∴ ∆G^{0} is positive.

(ii) The electrode reaction is written as,

2Ag ^{+}_{(aq)} + Cu_{(s)}→ Cu^{2 +}_{(aq)} + Ag_{(s)}

E^{0}_{cell} =

= + 0.80V - 0.34V

∴ E^{0}_{cell} = 0.46V

It is feasible, as E^{0}_{cell} is positive, ∴ ∆G^{0} is negative.

(iii) The electrode reaction is written as,

2Fe^{3 +}_{(aq)} + 2Br ^{-}_{(aq)}→ 2Fe^{2 +}_{(aq)} + Br_{2}

E^{0}_{cell} =

= 0.77V - 1.09V

∴ E^{0}_{cell} = - 0.32 V

It is not feasible, as E^{0}_{cell} is negative, ∴ ∆G^{0} is positive.

(iv) The electrode reaction is written as,

Ag_{(s)} + Fe^{3 +}_{(aq)} → Fe^{2 +}_{(aq)} + Ag ^{+}_{(aq)}

E^{0}_{cell} =

= 0.77V - 0.80V

∴ E^{0}_{cell} = - 0.03

It is not feasible, as E^{0}_{cell} is negative, ∴ ∆G^{0} is positive.

(v) The electrode reaction is written as,

Br_{2} + 2Fe^{2 +}_{(aq)} → 2Br ^{-}_{(aq)} + 2Fe^{3 +}_{(aq)}

E^{0}_{cell} =

= 1.09V - 0.77V

∴ E^{0}_{cell} = 0.32 V

It is feasible, as E^{0}_{cell} is positive, ∴ ∆G^{0} is negative.

**Question 18.**

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO_{3} with silver electrodes.

(ii) An aqueous solution of AgNO_{3} with platinum electrodes.

(iii) A dilute solution of H_{2}SO_{4} with platinum electrodes.

(iv) An aqueous solution of CuCl_{2} with platinum electrodes.

**Answer:**

Given -

All the ions are in aqueous state.

(i) Reaction in solution:

AgNO_{3(s)} + aq → Ag ^{+} + NO^{3 -}

H_{2}O Ã³ H ^{+} + OH ^{-}

At cathode:

Ag ^{+}_{(aq)} + e ^{-} →Ag_{(s)}

Ag ^{+} ions have lower discharge potential than H ^{+} ions. Hence, Ag ^{+} ions get deposited as Ag in preference to H ^{+} ions.

At anode:

Ag_{(s)}→ Ag ^{+}_{(aq)} + e ^{-}

As Ag anode is attacked by NO^{3 -} ions, Ag of the anode will dissolve to form Ag ^{+} ions in the aqueous solution.

(ii) Reaction in solution:

AgNO_{3(s)} + aq → Ag ^{+} + NO^{3 -}

H_{2}O Ã³H ^{+} + OH ^{-}

At cathode:

2Ag ^{+}_{(aq)} + 2e ^{-} →2Ag_{(s)}

Ag ^{+} ions have lower discharge potential than H ^{+} ions. Hence, Ag ^{+} ions get deposited as Ag in preference to H ^{+} ions.

At anode:

2OH ^{-}_{(aq)} → O_{2(g)} + 2H ^{+}_{(aq)} + 4e ^{-}

As anode is not attackable, out of OH ^{-} and NO^{3 -} ions, OH ^{-} having lower discharge potential, will be discharged in preference to NO^{3 -} ions. These then decompose to give out O_{2}.

(iii) Reaction in solution:

H_{2}SO_{4(aq)} → 2H ^{+}_{(aq)} + SO_{4}^{2 -}_{(aq)}

At cathode:

2H ^{+}_{(aq)} + 2e ^{-} →H_{2(g)}

At anode:

2OH ^{-}_{(aq)} → O_{2(g)} + 2H ^{+}_{(aq)} + 4e ^{-}

__∴____H _{2} gas is evolved at cathode and O_{2(g)} is evolved at anode.__

(iv) Reaction in solution:

CuCl_{2(s)} + aq → Cu^{2 +}_{(aq)} + Cl ^{-}_{(aq)}

H_{2}O Ã³H ^{+} + OH ^{-}

At cathode:

Cu^{2 +}_{(aq)} + 2e ^{-} →Cu_{(g)}

At anode:

2Cl ^{-}_{(aq)} - 2e ^{-} → Cl_{2(g)}

__∴____Cu will be deposited at cathode and Cl _{2} gas will be liberated at anode.__