Question

\text{The hydrogen electrode is dipped in a solution of pH = 3 at } 25^{\circ}\text{C. The potential of the electrode will be } -\\_\\_\\_\\_\\_ \times 10^{-2} \text{ V.}
$\left( \frac{2.303RT}{F} = 0.059 \, \text{V} \right)$

Answer 1

The potential of a hydrogen electrode in a solution can be calculated using the Nernst equation:

$E = E^0 - \frac{0.059}{n} \log \frac{1}{[\text{H}^+]}$

Given:
- $E^0 = 0$ (for the standard hydrogen electrode)
- pH = 3, so $[\text{H}^+] = 10^{-3} \, \text{M}$

Substitute into the equation:

$E = 0 - 0.059 \times \log(10^3) = -0.059 \times 3 = -0.177 \, \text{V}$

Thus, the electrode potential is:

$E = -17.7 \times 10^{-2} \, \text{V}$

The Correct answer is: 18