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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

The magnitude of the change in oxidising power of the \(MnO _{4}^{-} / Mn ^{2+}\) couple is \(x \times 10^{-4} \,V\), if the \(H ^{+}\) concentration is decreased from \(1\, M\) to \(10^{-4} \,M\) at \(25^{\circ} C\). (Assume concentration of \(MnO _{4}^{-}\) and \(Mn ^{2+}\) to be same on change in \(H ^{+}\) concentration). The value of \(x\) is ....... . (Rounded off to the nearest integer) \(\left[\right.\) Given \(\left.: \frac{2.303 RT }{ F }=0.059\right]\)

  1. A \(3776\)
  2. B \(3800\)
  3. C \(4276\)
  4. D \(1552\)
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Answer & Solution

Correct Answer

(A) \(3776\)

Step-by-step Solution

Detailed explanation

Eqn is- \(MnO _{4}^{-}+ H ^{\oplus}+5 e ^{-} \rightarrow Mn ^{+2}+4 H _{2} O\) Nernst equation: \(E _{\text {cell }}= E _{ Cell }^{0}-\frac{0.059}{5} \log \frac{\left[ Mn ^{+2}\right]}{\left[ MnO _{4}^{-}\right]}\left[\frac{1}{ H ^{+}}\right]^{8}\) \((I)\) Given…
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