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
On charging the lead storage battery, the oxidation state of lead changes from $\mathrm{x}_{1}$ to $\mathrm{y}_{1}$ at the anode and from $\mathrm{x}_{2}$ to $\mathrm{y}_{2}$ at the cathode. The values of $\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{x}_{2}, \mathrm{y}_{2}$ are respectively:
Match List-I with List-II: List-I