3.5     Calculate the emf of the cell in which the following reaction takes place:

              Ni(s)+2Ag^{+} (0.002M)\rightarrow Ni^{2+}(0.160M)+ 2Ag(s)

              Given that E^{\Theta }_{(cell) }= 1.05 \, V  

Answers (1)

Here we can directly apply the nernst equation :- 

                                     E_{Cell} = E_{cell}^{\circ}\ - \frac{0.0591}{n}log \frac{[Ni^{+2}]}{\left [ Ag^+ \right ]^2}

Putting the value in this equation :-  

                                      = 1.05\ - \frac{0.0591}{2}log \frac{0.160}{(0.002)^2}

or                                   = 1.05\ - 0.02955\ log (4\times10^4)

or                                    = 0.914\ V

Hence the required potential is 0.914 V.

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