For the following reaction 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂; E⁰(MnO₄⁻/Mn²⁺) = +1.51 V and E⁰(CO₂/H₂C₂O₄) = -0.49 V. At 298 K, the equilibrium constant is:

For the following reaction 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂; E⁰(MnO₄⁻/Mn²⁺) = +1.51 V and… · ABC26PH0398

Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

For the following reaction

$\mathrm{2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2}$; $E^0(\mathrm{MnO_4^-/Mn^{2+}}) = +1.51$ V and $E^0(\mathrm{CO_2/H_2C_2O_4}) = -0.49$ V.

At 298 K, the equilibrium constant is:

(a)$10^{500}$

(b)$10^{338}$

(c)$10^{38}$

(d)$10^{833}$

Answer

Answer (as printed): B

Explanation

No explanation was printed for this question.

Study loop for Electrochemistry

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app