If $\mathrm{KMnO}_{4}$ is reduced by oxalic acid in an acidic medium then oxidation number of $\mathrm{Mn}$…
- 4 to 2
- 6 to 4
- $+7$ to $+2$
- 7 to 4
Solution
$\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} ightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
(O.S. of $\mathrm{Mn}$ changes form $+7$ to $+2$ ).
Asked in: JEE-TOPICTESTS-CHEMISTRY
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