• # question_answer 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) $0.023\text{ }V$

 From the given cell reaction and Nernst equation, ${{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[N{{i}^{2+}}]}{{{[A{{g}^{+}}]}^{2}}}$ $=1.05V-\frac{0.0591}{2}\log \frac{[0.160]}{{{[0.002]}^{2}}}$ $=1.05-\frac{0.0591}{2}\log {{\left( 4\times 10 \right)}^{4}}$ $=1.05-0.14=0.19V$ ${{E}_{cell}}=0.91V$