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 \(\left.: \frac{2.303 RT }{ F }=0.059\right]\)
- A \(3776\)
- B \(3800\)
- C \(4276\)
- D \(1552\)
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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