JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the following reaction \(MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O , E ^{\circ}=1.51 V\) The quantity of electricity required in Faraday to reduce five moles of \(MnO _{4}^{-}\) is ..... .
- A \(35\)
- B \(25\)
- C \(12\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
The reaction given is: \(MnO _4^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _2 O\) In this reaction, \(1\, mole\) \(MnO _4{ }^{-}\)reduces to form \(Mn ^{2+}\) and for this reaction \(5 F e^{-}\)are required. \(MnO _4{ }^{-} \rightarrow Mn ^{2+}\) \(+7 \quad\quad+2\) Now,…
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