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Electronic Configuration and Oxidation States

What volume o...

Question

What volume of decimolar $K M n O_{4}$ solution will be needed for complete oxidation of $100 m l$ , semimilar oxalic acid and solution.

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Solution

${5 C}^{(-)} + {M n O}_{4}^{-} \to {M n}^{+ 2}$
$C_{2} O_{4}^{- 2} \to {2 C O}_{2} + 2 e^{(-)}$
Hence $N_{1} V_{1} = N_{2} V_{2}$
$(0.1) \times V_{1} \times 5 = 100 \times 2 \times 0.5$
$V_{1} = 200 m l$ of ${K M n O}_{4}$ .

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