CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

Question : 7 of 37

Marks: +1, -0

How much charge in terms of Faraday is required to reduce one mol of

{\MnO}_4^{-}

{\Mn}^{2+}

Solution:

4.825 × 10^5 {\C}

{\MnO}_4^{-}+5 {\e}^{-} → {\Mn} ↙{+7}^{2+}

\;\text " Charge "\;=5 × {\F}=5 × 96500 {\C}

=4.825 × 10^5 {\C}

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