Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s) + 2Ag⁺(0.001M) → Ni²⁺(0.001M) + 2Ag(s)
(Given that E°cell = 1.05 V, 2.303RT/F = 0.059 at 298 K)

Option 1 - 1.0385 V

Option 2 - 1.385 V

Option 3 - 0.9615 V

Option 4 - 1.05 V

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Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 3

Detailed Solution:

E_cell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 * 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? _cell is incorrectly given as 10.5 V. This should have been 1.05 volt.

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