Q. In acid medium, solution of potassium permanganate (mol. Wt. $158$) reacts with reducing agents according to the equation. $MnO _{-}^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O$. The mass of $KMnO _{4}$ required to prepare $500\, ml$ of $0.2\, N$ solution is

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$3.16$ grams

$15.8$ grams

$6.32$ grams

$31.6$ grams

Solution:

Correct answer is (a) $3.16$ grams

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6. Given below are half cell reactions:
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$ E _{ Mn ^{2+} / MnO _{4}^{-}}^{\circ}=-1.510\, V$
$\frac{1}{2} O _{2}+2 H ^{+}+2 e ^{-} \rightarrow H _{2} O ,$
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