Question

For an electrochemical cell, $Sn\left(s\right)\left|{\left(Sn\right)}^{2+}\left(aq,1M\right)\right|\left|{\left(Pb\right)}^{2+}\left(aq,1M\right)\right|Pb\left(s\right)$ , the ratio $\frac{\left[S{n}^{2+}\right]}{\left[P{b}^{2+}\right]}$ , when this cell attains equilibrium is
$Given:E_{S{n}^{2+}|Sn}^0=-0.14V{E}_{P{b}^{2+}|Pb}^0=-0.13V\frac{2.303RT}{F}=0.06$

• A. $4.3$
• B. $1$
• C. $-2.15$
• D. $2.14$

Answer 1

Answer: (D)

At equilibrium
$E_{\text{cell}}=0$
$E_{\text{cell}}^0=0.01\text{ V}$
$\text{Sn}+\text{P}{\text{b}}^{\text{2+}}\to \text{S}{\text{n}}^{\text{2+}}+\text{Pb}$
$E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.06}{n}logQ$
$0=0.01-\frac{0.06}{2}log\frac{\left[\text{S}{\text{n}}^{2+}\right]}{\left[\text{P}{\text{b}}^{2+}\right]}$
$0.01=\frac{0.06}{2}log\frac{\left[\text{S}{\text{n}}^{2+}\right]}{\left[\text{P}{\text{b}}^{2+}\right]}$
$\frac{1}{3}=log\frac{\left[\text{S}{\text{n}}^{2+}\right]}{\left[\text{P}{\text{b}}^{2+}\right]}$
$\frac{\left[{\text{Sn}}^{2+}\right]}{\left[{\text{Pb}}^{2+}\right]}=10^{\frac{1}{3}}=2.14$