Redox Reactions and Electrochemistry — CBSE Class 12 · 2018 Board Question with Solution

(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:

Sn (s) ∣ S n^{2 +} (0.004 M) ∣∣ H^{+} (0.020 M) ∣ H_{2} (g) (1 bar) ∣ Pt (s)

(Given:

E_{S n^{2 +} /Sn}^{\circ} = - 0.14

V)
(b) Give reasons:
(i) On the basis of E° values, O

_{2}

gas should be liberated at the anode, but it is Cl

_{2}

gas which is liberated in the electrolysis of aqueous NaCl.
(ii) The conductivity of CH

_{3}

COOH decreases on dilution.

Solution

(a) Cell reaction:

Sn (s) + 2 H^{+} (aq) ⟶ S n^{2 +} (aq) + H_{2} (g)

E_{ce l l}^{\circ} = 0.00 - (- 0.14) = 0.14 V, n = 2

Nernst equation:

E_{ce l l} = 0.14 - \frac{0.059}{2} lo g \frac{[ S n ^{2 +} ]}{[ H ^{+} ] ^{2}} = 0.14 - 0.0295 lo g \frac{0.004}{( 0.02 ) ^{2}}

= 0.14 - 0.0295 lo g 10 = 0.14 - 0.0295

E_{ce l l} = 0.11 V

(b)(i) Although E° favours oxidation of water to O

_{2}

(+1.23 V) over Cl

^{-}

to Cl

_{2}

(+1.36 V), the evolution of O

_{2}

at most anodes requires a high overpotential; the effective potential for O

_{2}

exceeds that for Cl

_{2}

, so Cl

^{-}

is discharged and Cl

_{2}

is liberated.
(ii) Conductivity counts ions in a unit volume. On dilution of CH

_{3}

COOH, although dissociation increases, the total number of ions per unit volume still falls — so κ decreases.

CBSE Class 12 Board · 2018 · Subjective