CBSE Class 12 · 2016 · All India · Redox Reactions and ElectrochemistryLong Answer

(a) Calculate

E_{ce l l}^{\circ}

for the following reaction at 298 K:

2Cr (s) + 3F e^{2 +} (0.01 M) ⟶ 2C r^{3 +} (0.01 M) + 3Fe (s)

Given:

E_{ce l l} = 0.261

V.
(b) Using the E° values of A and B, predict which one is better for coating the surface of iron [E°(Fe

^{2 +}

/Fe) = −0.44 V] to prevent corrosion, and why.
Given: E°(A

^{2 +}

/A) = −2.37 V; E°(B

^{2 +}

/B) = −0.14 V.

Solution

(a) Nernst equation (

n = 6

E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{6} lo g \frac{[ C r ^{3 +} ] ^{2}}{[ F e ^{2 +} ] ^{3}}

0.261 = E_{ce l l}^{\circ} - \frac{0.059}{6} lo g \frac{1 0 ^{- 4}}{1 0 ^{- 6}} = E_{ce l l}^{\circ} - \frac{0.059}{6} \times 2

E_{ce l l}^{\circ} = 0.261 + 0.0197

E_{ce l l}^{\circ} = 0.28 V

(b) Metal A (

E^{\circ} = - 2.37

V, like Mg) is better. Being more reactive than iron, A acts as the sacrificial anode and is oxidised preferentially even where the coating is broken.
B (

E^{\circ} = - 0.14

V) is less reactive than iron; once a B-coating is scratched, iron becomes the anode of the local cell and rusts even faster.

CBSE Class 12 Board · 2016 · All India · Subjective

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CBSE Class 12 · 2016 · All India · Redox Reactions and ElectrochemistryLong Answer

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