Question

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

Answer 1

Answer: 2.13

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