A) 0.77V
B) -0.829V
C) -0.77V
D) None of these
Correct Answer: C
Solution :
Cell reaction is \[2Ag(s)+2{{H}^{+}}(aq)\to 2A{{g}^{+}}(aq)+{{H}_{2}}(g)\] \[{{E}_{cell}}=-0.80-\frac{0.0591}{2}\log \frac{{{(0.1)}^{2}}\times 0.1}{{{(0.1)}^{2}}}\] \[{{E}_{cell}}=-0.80+0.02955=-0.77\text{V}\]
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