Question

For an electrochemical cell $Sn\left(s\right)|Sn^{2}+ \left(aq, 1M\right)||Pb^{2+} \left(aq, 1M\right)|Pb\left(s\right)$
the ratio $\frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]}$ when this cell attains equilibrium is __________ .
( Given : $E^{0}_{Sn^{2+}|Sn}=-0.14\,V, E^{0}_{Pb^{2+}|Pb}=-0.13\,V, \frac{2.303RT}{F}=0.06$ )
Given 1.07

Answer 1

Cell reaction is :
$\ce{Sn(s) + Pb^{+2}(aq)->Sn^{+2}(aq) + Pb(s)}$
Apply Nernst equation :
$E_{cell}=E^{0}_{cell}-\frac{0.06}{2}log \frac{\left[Sn^{+2}\right]}{\left[Pb^{+2}\right]}...\left(1\right)$
$E^{0}_{cell}=-0.13 + 0.14 = 0.01\, V$
At equilibrium : $E_{cell} = 0$
Substituting in \left(1\right)
$0=0.01-\frac{0.06}{2}log \frac{\left[Sn^{+2}\right]}{\left[Pb^{+2}\right]}$
$\Rightarrow \frac{1}{3}=log \frac{\left[Sn^{+2}\right]}{\left[Pb^{+2}\right]}$
$\Rightarrow \frac{\left[Sn^{+2}\right]}{\left[Pb^{+2}\right]}=2.15$