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CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

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Question : 7 of 37
Marks: +1, -0
How much charge in terms of Faraday is required to reduce one mol of MnO4−\mathrm{MnO}_4^{-}to Mn2+\mathrm{Mn}^{2+} ?
Solution:  
4.825×105 C4.825 \times 10^5 \, \mathrm{C}
MnO4−+5 e−→Mn+72+\mathrm{MnO}_4^{-} + 5 \, \mathrm{e}^{-} \rightarrow \underset{+7}{\mathrm{Mn}}^{2+}
   Charge   =5×F=5×96500 C\; \text{ Charge } \; = 5 \times F = 5 \times 96500 \, \mathrm{C} =4.825×105 C= 4.825 \times 10^5 \, \mathrm{C}
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