What is the emf of the cell at $298\,K$ in which the following reaction takes place?
\[
Ni(s) + 2Ag^+(0.002M) \rightarrow Ni^{2+}(0.04M) + 2Ag(s)
\]
($E^\circ_{cell} = 1.05V$)
Show Hint
- 1.16 V
- 0.93 V
- 0.73 V
- 0.83 V
- 1.32 V
The Correct Option is B
Solution and Explanation
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