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

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

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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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