Chemistry - Electrochemistry Question with Solution | TestHub

$E_{Cu|C{u}^{2+}}=E_{Cu|C{u}^{2+}}^{\circ }-\frac{0.059}{2}\log [C{u}^{2+}]$ …(i)

$E_{Cu|C{u}^{2+}}^{\prime }=E_{Cu|C{u}^{2+}}^{\circ }-\frac{0.059}{2}\log \frac{[C{u}^{2+}]}{100}$ …(ii)

Subtracting (i) from (ii) gives:

$E_{Cu|C{u}^{+2^{}}}-E_{Cu|C{u}^{+2}}^{\prime }=\frac{0.059}{2}\left[\log \frac{[C{u}^{2+}]}{100}-\log [C{u}^{2+}]\right]$

$\Delta E_{Cu|C{u}^{+2}}=\frac{0.059}{2}\log 10^{-2}=-0.059;\Delta E_{Cu|C{u}^{+2}}=-0.059V$