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

Correct Answer: 18

The potential of a hydrogen electrode in a solution is determined by the Nernst equation. For the hydrogen electrode, the reaction is:
\(\text{H}_2(g) \rightleftharpoons 2\text{H}^+(aq) + 2e^-\) 
The Nernst equation is:
\(E = E^{\circ} - \frac{2.303RT}{nF} \log \left( \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \right)\)
For a standard hydrogen electrode, \(E^{\circ} = 0\) V, the pressure of \(\text{H}_2\) gas is 1 atm, and \(n = 2\). At a given \text{pH}, \([\text{H}^+]=10^{-\text{pH}}\). Therefore, the equation simplifies to:
\(E = -\frac{2.303RT}{F} \frac{\log (10^{-\text{pH}})^2}{2}\)
This further simplifies to:
\(E = -\frac{0.059}{1} \ \text{V} \times \text{pH}\)
Given \(\text{pH} = 3\), substitute into the equation:
\(E = -0.059 \times 3 \ \text{V} \)
Thus,
\(E = -0.177 \ \text{V}\)
Expressed as \(-17.7 \times 10^{-2} \ \text{V}\), this value confirms it is within the specified range of 18,18 when considering absolute value: \(17.7 \approx 18\).
Therefore, the potential of the electrode is:
\(-17.7 \times 10^{-2} \ \text{V}\).

Answer 2

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