i) Explain why electrolysis of an aqueous solution of NaCl gives H$_2$ at the cathode and Cl$_2$ at the anode. Write the overall reaction. ($E^{\circ}_{\mathrm{Na^+/Na}} = -2.71$ V; $E^{\circ}_{\mathrm{H_2O/H_2}} = -0.83$ V; $E^{\circ}_{\mathrm{Cl_2/2Cl^-}} = +1.36$ V; $E^{\circ}_{\mathrm{H^+,O_2/H_2O}} = +1.23$ V) ii) Calculate the emf of the cell $$\mathrm{Zn\,|\,Zn^{2+}(0.1\ M)\,||\,Cd^{2+}(0.01\ M)\,|\,Cd}$$ at 298 K. (Given: $E^{\circ}_{\mathrm{Zn^{2+}/Zn}} = -0.76$ V, $E^{\circ}_{\mathrm{Cd^{2+}/Cd}} = -0.40$ V)

(i) Cathode: water (−0.83 V) reduced before Na⁺ (−2.71 V) → H₂; anode: Cl⁻ oxidised (O₂ overpotential) → Cl₂. Overall: 2NaCl+2H₂O → 2NaOH+H₂+Cl₂. (ii) E_cell ≈ 0.33 V.

CBSE Class 12 · 2006 · Delhi · Redox Reactions and ElectrochemistryLong Answer

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i) Explain why electrolysis of an aqueous solution of NaCl gives H

_{2}

at the cathode and Cl

_{2}

at the anode. Write the overall reaction.
(

E_{N a^{+} /Na}^{\circ} = - 2.71

E_{H_{2} O/ H_{2}}^{\circ} = - 0.83

E_{C l_{2} /2C l^{-}}^{\circ} = + 1.36

E_{H^{+}, O_{2} / H_{2} O}^{\circ} = + 1.23

V)
ii) Calculate the emf of the cell

Zn ∣ Z n^{2 +} (0.1 M) ∣∣ C d^{2 +} (0.01 M) ∣ Cd

at 298 K.
(Given:

E_{Z n^{2 +} /Zn}^{\circ} = - 0.76

E_{C d^{2 +} /Cd}^{\circ} = - 0.40

Solution

(i) Cathode: of Na

^{+}

(−2.71 V) and water (−0.83 V), water has the higher reduction potential, so it is reduced:

2 H_{2} O + 2 e^{-} ⟶ H_{2} + 2O H^{-}

Anode: although E° favours O

_{2}

(+1.23 V) over Cl

_{2}

(+1.36 V), the high overpotential of oxygen means Cl

^{-}

is actually discharged:

2C l^{-} ⟶ C l_{2} + 2 e^{-}

Overall:

2NaCl (aq) + 2 H_{2} O (l) ⟶ 2NaOH (aq) + H_{2} (g) + C l_{2} (g)

(ii)

E_{ce l l}^{\circ} = - 0.40 - (- 0.76) = 0.36 V, n = 2

E_{ce l l} = 0.36 - \frac{0.059}{2} lo g \frac{0.1}{0.01} = 0.36 - 0.0295

E_{ce l l} = 0.33 V

CBSE Class 12 Board · 2006 · Delhi · Subjective

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CBSE Class 12 · 2006 · Delhi · Redox Reactions and ElectrochemistryLong Answer

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