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CBSE Class 12 · 2007 · Delhi · Redox Reactions and ElectrochemistryLong Answer

Calculate the emf of the cell:

Mg (s) ∣ M g^{2 +} (0.1 M) ∣∣ C u^{2 +} (1 \times 1 0^{- 3} M) ∣ Cu (s)

(Given:

E_{C u^{2 +} /Cu}^{\circ} = + 0.34

V,

E_{M g^{2 +} /Mg}^{\circ} = - 2.37

V)

Answer: E_cell ≈ 2.65 V

Solution

E_{ce l l}^{\circ} = 0.34 - (- 2.37) = 2.71 V, n = 2

Nernst equation:

E_{ce l l} = 2.71 - \frac{0.059}{2} lo g \frac{0.1}{1 0 ^{- 3}} = 2.71 - 0.0295 \times 2

= 2.71 - 0.059

E_{ce l l} = 2.65 V

CBSE Class 12 Board · 2007 · Delhi · Subjective

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