Published by:
CGP EDU Academic Team
Published on: August 11, 2026
Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s) + 2Ag + (0.001 M) 
Ni 2+ (0.001 M) +2 Ag(s)
(Given that E° cell =1.05 V, 
= 0.059 at 298 K)
(Please note that E° cell =10.5 V is modified to E° cell = 1.05V for accuracy point of view)
Text Solution
Verified by ExpertsThe correct answer is:
NOT AVAILABLE
Ni(s) + 2Ag + (0.001 M) 
Ni 2+ (0.001 M) + 2Ag(s)
E° cell = 1-05V
= 1.05 – 0.0295 x 3
= 1.05 – 0.0885
= 0.9615V
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