The magnitude of the change in oxidising power of the couple is x × 10 V, if the H concentration is — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the half-reaction and Nernst equation** For the MnO₄⁻/Mn²⁺ couple: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Use Nernst equation: E = E° + (0.0592/n) log Q **Step 2: Write the reaction quotient** Q = [MnO₄⁻][H⁺]⁸/[Mn²⁺] Since [MnO₄⁻] and [Mn²⁺] are equal and constant, they cancel in the ratio. Q ∝ [H⁺]⁸ **Step 3: Calculate E₁ at [H⁺] = 1 M** E₁ = E° + (0.0592/5) log(1)⁸ E₁ = E° + 0 = E° **Step 4: Calculate E₂ at [H⁺] = 10⁻⁴ M** (decreased from 1 M to 10⁻⁴ M) E₂ = E° + (0.0592/5) log(10⁻⁴)⁸ E₂ = E° + (0.01184) log(10⁻³²) E₂ = E° + (0.01184) × (-32) E₂ = E° - 0.3789 V **Step 5: Calculate change in oxidizing power** ΔE = E₁ - E₂ = E° - (E° - 0.3789) ΔE = 0.3789 V = 3.789 × 10⁻¹ V **Step 6: Express in scientific notation** ΔE = 3.789 × 10⁻¹ = 3776 × 10⁻⁴ V