Question

The reduction potential of hydrogen half-cell will be negative if

• A. \( \text{p}(\text{H}_2) = 1 \, \text{bar} \, \text{and} \, [\text{H}] = 1 \, \text{M} \)
• B. \( \text{p}(\text{H}_2) = 1 \, \text{bar} \, \text{and} \, [\text{H}] = 2 \, \text{M} \)
• C. \( \text{p}(\text{H}_2) = 2 \, \text{bar} \, \text{and} \, [\text{H}] = 1 \, \text{M} \)
• D. \( \text{p}(\text{H}_2) = 2 \, \text{bar} \, \text{and} \, [\text{H}] = 2 \, \text{M} \)

Hint:

The reduction potential is determined by the concentrations of Hand H2, as described by the Nernst equation.

Answer 1

The Correct Option is C

The reduction potential is influenced by the pressure of $H_2$ and the concentration of Hions. According to the Nernst equation, when the pressure is increased to 2 bar while the concentration of [H] is 1 M, the reduction potential becomes negative.