KCET · Chemistry · Electrochemistry
The charge required for the reduction of \( 1 \) mole of \( \mathrm{MnO}_{4}{ }^{-} \)to \( \mathrm{MnO}_{2} \) is
- A \( 1 \mathrm{~F} \)
- B \( 3 \mathrm{~F} \)
- C \( 5 \mathrm{~F} \)
- D 7 F
Answer & Solution
Correct Answer
(B) \( 3 \mathrm{~F} \)
Step-by-step Solution
Detailed explanation
According to the reaction
\( \mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 e^{-} \rightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \)
\( 3 \mathrm{~F} \) of charge required to convert \( 1 \) mole of \( \mathrm{MnO}_{4}{ }^{-} \)to \( \mathrm{MnO}_{2} \)
\( \mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 e^{-} \rightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \)
\( 3 \mathrm{~F} \) of charge required to convert \( 1 \) mole of \( \mathrm{MnO}_{4}{ }^{-} \)to \( \mathrm{MnO}_{2} \)
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