Question

The slopes of reduction potential vs. pH plots for the reactions: 1. \(NiO + 2H^+ + 2e^- \rightleftharpoons Ni + H_2O\) 2. \(2H^+ + 2e^- \rightleftharpoons H_2\) at 298 K are \(S_1\) and \(S_2\). The ratio \(\dfrac{S_1}{S_2}\) is (rounded off to one decimal place).

Hint:

Always check the ratio \(m/n\) (protons/electrons). If two reactions have the same \(m/n\), their slopes vs pH are identical.

Answer 1

Correct Answer: 0.8

Step 1: General Nernst slope.
For a reaction involving \(m\) protons and \(n\) electrons: \[ E = E^\circ - \frac{0.0591}{n} \, m \, pH \] Hence, slope = \(-\frac{0.0591 m}{n}\).

Step 2: First reaction.
\(NiO + 2H^+ + 2e^- \to Ni + H_2O\) Here, \(m=2, n=2\). \[ S_1 = -\frac{0.0591 \times 2}{2} = -0.0591 \]

Step 3: Second reaction.
\(2H^+ + 2e^- \to H_2\) Here also, \(m=2, n=2\). \[ S_2 = -\frac{0.0591 \times 2}{2} = -0.0591 \]

Step 4: Ratio.
\[ \frac{S_1}{S_2} = \frac{-0.0591}{-0.0591} = 1.0 \] Final Answer:
\[ \boxed{1.0} \]