Question

The slopes of reduction potential versus pH plots for the two reactions, \( {NiO} + 2{H}^+ + 2e^- \Rightarrow {Ni} + {H}_2{O} \) and \( 2{H}^+ + 2e^- \Rightarrow {H}_2 \), at 298 K and one atmospheric pressure are \( S_1 \) and \( S_2 \), respectively. The ratio \( \frac{S_1}{S_2} \) is (rounded off to one decimal place) ............

Hint:

For reduction potential versus pH plots, the slope is related to the number of electrons involved in the reaction. A higher number of electrons leads to a smaller slope.

Answer 1

The slope of the reduction potential versus pH plot for a half-reaction is given by the equation:
\[ \frac{dE}{dpH} = \frac{0.0592}{n} \] where \( n \) is the number of electrons involved in the reaction. For the given reactions:
- The first reaction \( {NiO} + 2{H}^+ + 2e^- \Rightarrow {Ni} + {H}_2{O} \) involves 2 electrons, so the slope \( S_1 = \frac{0.0592}{2} = 0.0296 \, {V/pH} \).
- The second reaction \( 2{H}^+ + 2e^- \Rightarrow {H}_2 \) also involves 2 electrons, so the slope \( S_2 = \frac{0.0592}{2} = 0.0296 \, {V/pH} \).
Since the slopes are equal, the ratio \( \frac{S_1}{S_2} = 1.0 \).