A) 0.73 V
B) 0.91 V
C) 0.62 V
D) 0.34 V
Correct Answer: B
Solution :
From the given cell reaction and Nemst equation, \[{{E}_{Cell}}=E_{Cell}^{\circ }-\frac{0.0591}{n}\log \frac{[N{{i}^{2+}}]}{{{[A{{g}^{+}}]}^{2}}}\] \[=1.05V-\frac{0.0591}{2}\log \,\,\,\,\frac{[0.160]}{{{[0.002]}^{2}}}\] \[=1.05-\frac{0.0591}{2}\log \,\,(4\times {{10}^{4}})=1.05-\frac{0.0591}{2}(4.6021)\]\[=1.05-0.14=+0.91V\] \[{{E}_{Cell}}=+0.91V\]
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