Question

For the cell reaction $Pb+S{n}^{2+}\to P{b}^{2+}+Sn$
Calculate the ratio of cation concentration for which E = 0. Given an example where $E^o$ is zero.
Given that $Pb\to P{b}^{2+}$; $E^o$ = 0.13V
$S{n}^{2+}+2e^{-}\to Sn$; $E^o$ = -0.14V

Answer 1

$Pb+S{n}^{2+}⟶P{b}^{2+}+Sn$

Cell reaction= $Sn+P{b}^{2+}⟶Pb+S{n}^{2+}$

$E_{red}^o$ for $P{b}^{2+}⟶Pb=-0.13V$

$E_{red}^o$ for $S{n}^{2+}⟶Sn=-0.14V$

Thus $P{b}^{2+}$ has high reduction potential, thus it will be at cathode & $S{n}^{+2}$ will act as anode.

$E_{cell}^o=(E_{red}^o{)}_c-(E_{red}^o{)}_A$

$\Rightarrow E_{cell}^o=-0.13V-(-0.14V)$

$=0.01V$

$\Rightarrow E_{cell}=E^{o}cell-\frac{0.0591}{2}\log \frac{S{n}^{2-1}}{P{b}^{2+}}$

Let put $E_{cell}=0$

$\Rightarrow 0=0.01-\frac{0.0591}{2}\log \frac{S{n}^{2+}}{P{b}^{2+}}$

$\Rightarrow \frac{0.01\times 2}{0.0591}=\log \frac{S{n}^{2+}}{P{b}^{2+}}$

$\Rightarrow 0.338=\log \frac{S{n}^{2+}}{P{b}^{2+}}$

$\Rightarrow {10}^{0.338}=\frac{S{n}^{2+}}{P{b}^{2+}}$

$\Rightarrow \frac{S{n}^{2+}}{P{b}^{2+}}=2.179$

Thus the ratio of cations concentration is $2.179$ .