Question

For the cell reaction:

$Pb+S{n}^{2+}\to P{b}^{2+}+Sn$

Given that: $Pb\to P{b}^{2+},E^o=0.13V$

$S{n}^{2+}+2e^{-}\to Sn;E^o=-0.14V$

What would be the ratio of cation concentration for which E = 0?

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

Answer: (B)

$\frac{1}{2}$

For the given cell reaction,

$E_{cell}^o=-0.14V+0.13V=-0.01V$

According to Nernst equation:

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}log\frac{[P{b}^{2+]}}{\left[S{n}^{2+}\right]}$

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

$\Rightarrow log\frac{\left[P{b}^{2+}\right]}{\left[S{n}^{2+}\right]}=\frac{0.01}{-0.0296}=-0.3$

$\Rightarrow \frac{\left[P{b}^{2+}\right]}{\left[S{n}^{2+}\right]}=antilog(-0.3)$

$\Rightarrow \frac{\left[P{b}^{2+}\right]}{\left[S{n}^{2+}\right]}=0.5=\frac{1}{2}$

Hence, option B is correct.