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  • Give below the equation <img alt="\begin{gathered} \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\ \mathrm{E}^{\circ}=1.51 \mathrm{v} \e

Give below the equation
\begin{gathered} \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\ \mathrm{E}^{\circ}=1.51 \mathrm{v} \end{gathered}
The quantity of electricity required in faraday to reduce 6 moles of \mathrm{MnO}_4^{-} is.

Option: 1

20 faraday


Option: 2

30 faraday


Option: 3

25 faraday


Option: 4

6 faraday


Answers (1)

best_answer

In the reduction half cell
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \rightarrow 5 \mathrm{Mn}^{2+}+4 \mathrm{H} \\
1 \mathrm{~mol} \rightarrow 5 \mathrm{~F} \\

6 \text { mole } \rightarrow 30 \mathrm{~F} \\
As one mole of \mathrm{MnO}_4^{-} require 5 F charge
Q = 5 x 6 Faraday = 30 Faraday
 

Posted by

shivangi.bhatnagar

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