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CBSE Class 12 · 2024 · Redox Reactions and ElectrochemistryShort Answer

Calculate the emf of the following cell:

Ni (s) + 2A g^{+} (0.01 M) ⟶ N i^{2 +} (0.1 M) + 2Ag (s)

Given:

E_{ce l l}^{\circ} = 1.05

V, log 10 = 1.

Answer: E_cell ≈ 0.96 V

Solution

Nernst equation (

n = 2

):

E_{ce l l} = 1.05 - \frac{0.059}{2} lo g \frac{[ N i ^{2 +} ]}{[ A g ^{+} ] ^{2}} = 1.05 - 0.0295 lo g \frac{0.1}{( 0.01 ) ^{2}}

= 1.05 - 0.0295 lo g 1 0^{3} = 1.05 - 0.0885

E_{ce l l} = 0.96 V

CBSE Class 12 Board · 2024 · Subjective

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