Question

For the cell
$Cu(s)\left|C{u}^{2+}(aq)(0.1M)\Vert A{g}^{+}(aq)(0.01M)\right|Ag(s)$
the cell potential $E_1=0.3095V$ For the cell
$Cu(s)\left|C{u}^{2+}(aq)(0.01M)\Vert A{g}^{+}(aq)(0.001M)\right|Ag(s)$
the cell potential =_____$10^{-2}V$.
(Round off the Nearest Integer).
[Use : $\frac{2.303RT}{F}=0.059$]

Answer 1

Answer: 28

Cell reaction is :
$Cu(s)+2A{g}^{+}(aq)\to C{u}^{2+}(aq)+2Ag(s)$
Now, $E_{\text{cell }}=E_{\text{Cell }}^{\circ }-\frac{0.059}{2}\log \frac{\left[C{u}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}......(1)$
$\therefore E_1=0.3095=E_{\text{Cell }}^{\circ }-\frac{0.059}{2}\cdot \log \frac{0.01}{(0.001{)}^2}$
From $(1)$ and $(2),E_2=0.28V=28\times 10^{-2}V$