| Calculate the emf of the cell in which the following reaction takes place. |
| \[Ni(s)+2A{{g}^{+}}(0.002M)\xrightarrow{{}}N{{i}^{2+}}(0.160M)+2Ag\,(s)\] Given that \[E_{cell}^{{}^\circ }=1.05V\] |
A) \[0.61\text{ }V\]
B) \[0.81\text{ }V\]
C) \[0.82\text{ }V\]
D) \[0.023\text{ }V\]
Correct Answer: B
Solution :
| From the given cell reaction and Nernst equation, |
| \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[N{{i}^{2+}}]}{{{[A{{g}^{+}}]}^{2}}}\] |
| \[=1.05V-\frac{0.0591}{2}\log \frac{[0.160]}{{{[0.002]}^{2}}}\] |
| \[=1.05-\frac{0.0591}{2}\log {{\left( 4\times 10 \right)}^{4}}\] |
| \[=1.05-0.14=0.19V\] |
| \[{{E}_{cell}}=0.91V\] |
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