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CBSE Class 12 · 2015 · Redox Reactions and ElectrochemistryShort Answer

Calculate the e.m.f. of the following cell at 298 K:

2Cr (s) + 3F e^{2 +} (0.1 M) ⟶ 2C r^{3 +} (0.01 M) + 3Fe (s)

Given:

E^{\circ} (C r^{3 +} ∣Cr) = - 0.74

V,

E^{\circ} (F e^{2 +} ∣Fe) = - 0.44

V.

Answer: E_cell ≈ 0.31 V

Solution

E_{ce l l}^{\circ} = - 0.44 - (- 0.74) = 0.30 V, n = 6

Nernst equation:

E_{ce l l} = 0.30 - \frac{0.059}{6} lo g \frac{[ C r ^{3 +} ] ^{2}}{[ F e ^{2 +} ] ^{3}} = 0.30 - \frac{0.059}{6} lo g \frac{1 0 ^{- 4}}{1 0 ^{- 3}}

= 0.30 + \frac{0.059}{6} = 0.30 + 0.0098

E_{ce l l} = 0.31 V

CBSE Class 12 Board · 2015 · Subjective

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