MHT CET · Chemistry · Electrochemistry
Calculate the amount of electricity required in coulombs to convert 0.08 mol of \(\mathrm{MnO}_4^{-}\)to \(\mathrm{Mn}^{2+}\).
- A 96500 C
- B 38600 C
- C 48250 C
- D 19300 C
Answer & Solution
Correct Answer
(B) 38600 C
Step-by-step Solution
Detailed explanation
The reaction is
\(\mathrm{MnO}_4^{-}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\)
For reduction of 1 mole, 5 F electricity is required
\(\therefore \quad\) For 0.08 mole, 0.4 F electricity is required.
\(\begin{aligned}
& 1 \mathrm{~F}=96500 \mathrm{C} \\
\therefore \quad & 0.4 \mathrm{~F}=96500 \times 0.4=38600 \mathrm{C}
\end{aligned}\)
\(\mathrm{MnO}_4^{-}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\)
For reduction of 1 mole, 5 F electricity is required
\(\therefore \quad\) For 0.08 mole, 0.4 F electricity is required.
\(\begin{aligned}
& 1 \mathrm{~F}=96500 \mathrm{C} \\
\therefore \quad & 0.4 \mathrm{~F}=96500 \times 0.4=38600 \mathrm{C}
\end{aligned}\)
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