Consider the following redox reaction, The standard reduction potentials are given as below

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Solution

Q.
Consider the following redox reaction:

${MnO}_{4}^{-} + H^{+} + H_{2} C_{2} O_{4} ⇌ {Mn}^{2 +} + H_{2} O + {CO}_{2}$

The standard reduction potentials are given as below $(E_{r e d}^{°}) :$

$E_{{MnO}_{4}^{-} / {Mn}^{2 +}}^{\circ} = + 1.51 V$

$E_{{CO}_{2} / H_{2} C_{2} O_{4}}^{\circ} = - 0.49 V$

If the equilibrium constant of the above reaction is given as $K_{eq} = 10^{x},$ then the value of $x$ = ________ (nearest integer) [2024]

Ans.
(338)

   $2 \times ({MnO}_{4}^{-} + 5 e^{-} + 8 H^{+} \to {Mn}^{2 +} + 4 H_{2} O), Δ G_{1}^{°} = - 10 F E_{{MnO}_{4}^{-} / {Mn}^{2 +}}^{\circ}$

$\frac{5 \times (H_{2} C_{2} O_{4} \to 2 {CO}_{2} + 2 H^{+} + 2 e^{-}), Δ G_{2}^{°} = - 10 F E_{H_{2} C_{2} O_{4} / {CO}_{2}}^{\circ}}{2 {MnO}_{4}^{-} + 5 H_{2} C_{2} O_{4} + 6 H^{+} \to 10 {CO}_{2} + 2 {Mn}^{2 +} + 8 H_{2} O, Δ G_{n e t}^{°} = Δ G_{2}^{°} + Δ G_{1}^{°}}$

$Δ G_{n e t}^{°} = Δ G_{2}^{°} + Δ G_{1}^{°}$

$- 2.303 R T \log_{10} K = - 10 F E_{{MnO}_{4}^{-} / {Mn}^{2 +}}^{\circ} - 10 F E_{H_{2} C_{2} O_{4} / {CO}_{2}}^{\circ}$

   $- 2.303 R T \log_{10} K = - 10 F (1.51) - 10 F (0.49) [E_{H_{2} C_{2} O_{4} / {CO}_{2}}^{\circ} = - E_{{CO}_{2} / H_{2} C_{2} O_{4}}^{\circ} = - (- 0.49 V)]$

   $\log_{10} K = \frac{20 F}{2.303 R T}$

$\log_{10} K = \frac{20 \times 96500}{2.303 \times 8.314 \times 298}$

   $\log_{10} K = 338.25$

   $K = 10^{338.25}$

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