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CBSE Class 12 · 2010 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

Calculate the emf of the cell in which the following reaction takes place:

Ni (s) + 2A g^{+} (0.002 M) ⟶ N i^{2 +} (0.160 M) + 2Ag (s)

Given:

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

V.

Answer: E_cell ≈ 0.91 V

Solution

Nernst equation (

n = 2

):

E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{2} lo g \frac{[ N i ^{2 +} ]}{[ A g ^{+} ] ^{2}}

Substituting:

E_{ce l l} = 1.05 - 0.0295 lo g \frac{0.160}{( 0.002 ) ^{2}} = 1.05 - 0.0295 lo g (4 \times 1 0^{4})

= 1.05 - 0.0295 \times 4.602 = 1.05 - 0.136

E_{ce l l} = 0.914 V

CBSE Class 12 Board · 2010 · Delhi Compartment · Subjective

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