Question

Standard cell voltage for the cell $Pb/P{b}^{2+}||S{n}^{2+}/Sn$ is $-0.01V$. if the cell is to exhibits $E_{cell}=0$, the value of $log\left[S{n}^{2+}/\right[P{b}^{2+}]$ should be:

• A. $0.33$
• B. $0.5$
• C. $1.5$
• D. $-0.5$

Answer 1

Answer: (A)

$0.33$

According to Nernst equation

$E_{cell}=E_{cell}^o-\frac{RT}{nF}ln\theta$ $[E_{cell}^o=$ Standard cell voltage$]$

$=E_{cell}^o-\frac{0.0591}{n}log\theta$

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

$\because Pb\left(s\right)\rightleftharpoons p{b}^{2+}\left(aq\right)+2e^{-}$ $($ Oxidation $)$.................$\left(i\right)$

$\because S{n}^{2+}\left(aq\right)+2e\rightleftharpoons Sn\left(s\right)$ $($ Reduction $)$.................$\left(ii\right)$

On combining equations $i$ and equation $ii$

$pb\left(s\right)+S{n}^{2+}\left(aq\right)\rightleftharpoons Sn+P{b}^{2+}\left(aq\right)$

$\therefore Q=\frac{\left[P{b}^{2+}\right]}{\left[S{n}^{2+}\right]}$

$n=2$

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

$\because E_{cell}=0$

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

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

$log\frac{\left[P{b}^{2+}\right]}{S{n}^{2+}]}=-0.33$

$log\frac{\left[S{n}^{2+}\right]}{\left[P{b}^{2+}\right]}=0.33$