Question

For an electrochemical cell, $Sn\left(s\right)\left|\left(Sn\right)^{2 +} \left(aq , 1 M\right)\right|\left|\left(Pb\right)^{2 +} \left(aq , 1 M\right)\right|Pb\left(s\right)$ , the ratio $\frac{\left[Sn^{2 +}\right]}{\left[Pb^{2 +}\right]}$ , when this cell attains equilibrium is
$Given:E_{Sn^{2 +} \left|\right. Sn}^{0}=-0.14VE_{Pb^{2 +} \left|\right. Pb}^{0}=-0.13V\frac{2 .303 RT}{F}=0.06$

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

Answer 1

Answer: (D)

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