Question

If $E_{\left(P{b}^{2+}|Pb\right)}^{\ominus }=-0.126V,E_{\left(Z{n}^{2+}|Zn\right)}^{\ominus }=-0.763V$ at ${25}^{\circ }C$, calculate EMF of the cell: $Zn|Z{n}^{2+}\left(1M\right)||P{b}^{2+}$
(1 M) | Pb. If an excess of metallic zn added to 1 M concentration of $P{b}^{2+}$ ions, The concentration of $P{b}^{2+}$ ions at equilibrium is $2.5\times {10}^{-x}M$
value of x is..............

Answer 1

$Zn+P{b}^{2+}\to Z{n}^{2+}+Pb$
$E_{cell}^0=-0.126-(-0.763)=0.637V$
$E_{cell}=E_{cell}^0-\frac{0.059}{2}log\frac{\left[Z{n}^{2+}\right]}{\left[P{b}^{2+}\right]}$
At equilibrium, $E_{cell}=o$
$\Rightarrow E_{cell}^{\ominus }=\frac{0.059}{2}log\frac{\left[Z{n}^{2+}\right]}{\left[P{b}^{2+}\right]}$
$\Rightarrow 0.637=\frac{0.059}{2}log\frac{1}{\left[P{b}^{2+}\right]}$
$\Rightarrow log\left[P{b}^{2+}\right]=\frac{-0.637\times 2}{0.059}$
$\Rightarrow \left[P{b}^{2+}\right]=2.5\times {10}^{-22}M=2.5\times {10}^{-x}M$
$x=22$