Question

What ratio of $P{b}^{2+}$ to $S{n}^{2+}$ concentration is needed to reverse the following cell reaction?
$Sn\left(s\right)+P{b}^2(aq.)\to S{n}^{2+}(aq.)+Pb\left(s\right)$
$E_{S{n}^{2+}/Sn}^{\circ }=-0.136volt$ and $E_{P{b}^{2+}/Pb}^{\circ }=-0.126volt$.

Answer 1

In the given cell reaction,

$E_{cell}$ = $E_{cell}^0$ $-$ $\frac{0.059}{2}$ $log$$\frac{S{n}^{2+}}{P{b}^{2+}}$

$E_{cell}$ = $E_{cell}^0$ $+$ $\frac{0.059}{2}$ $log$$\frac{P{b}^{2+}}{S{n}^{2+}}$

$E_{cell}$= 0.010 + 0.0295 $log$$\frac{P{b}^{2+}}{S{n}^{2+}}$

when $E_{cell}$ is negative then cell reaction reverses,

if $\frac{P{b}^{2+}}{S{n}^{2+}}$ is in between 0.1 to 0.9 then, cell reaction reverse.