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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

Consider the following redox reaction : \(\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\) The standard reduction potentials are given as below \(\left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)\) \(\mathrm{E}_{\mathrm{MmO}_4^{-} / \mathrm{Mm}^{2+}}^{\circ}=+1.51 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}\) If the equilibrium constant of the above reaction is given as \(K_{\text {eq }}=10^x\), then the value of \(x=\)_______. (nearest integer)

  1. A \(339\)
  2. B \(350\)
  3. C \(390\)
  4. D \(340\)
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Answer & Solution

Correct Answer

(A) \(339\)

Step-by-step Solution

Detailed explanation

Cell \(\mathrm{Rx}^{\mathrm{n}} ; \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2\)…
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