# What is the electrode potential for Pb^(2+)(aq) + 2e^(-) rightleftharpoons Pb(s) when the concentration of Pb^(2+) (aq) is 0.05 mol dm^(-3) ?

## Temperature is 298 K and the standard electrode potential for the half-cell reaction is $- 0.13$ V.

Dec 5, 2017

-0.16 V

#### Explanation:

The expression for the electrode potential of a half - cell is:

$\textsf{E = {E}^{\circ} - \frac{R T}{z F} \ln \left(\left[\left[\text{reduced form"]]/[["oxidised form}\right]\right]\right)}$

For this half - cell at 298K this can be simplified to:

$\textsf{E = {E}^{\circ} + \frac{0.0591}{2} \log \left[P {b}^{2 +}\right]}$

$\therefore$$\textsf{E = - 0.13 + \frac{0.051}{2} \log \left[0.05\right]}$

$\textsf{E = - 0.13 - 0.3317 = - 0.16 \textcolor{w h i t e}{x} V}$

We would expect the potential to become more -ve like this since Le Chatelier's Principle predicts that reducing $\textsf{\left[P {b}^{2 +}\right]}$ will cause the equilibrium to shift to the left thus pushing out more electrons.