Question #d2fbd

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Question #d2fbd

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Answer 1 · 2016-06-05T19:44:52

Manganate, $M n (V I)$ $Mn(VI)$ , is reduced to $M n (I I I)$ $Mn(III)$ . Oxalic acid is oxidized to $C O_{2}$ $CO_2$ .

Explanation:

Reduction half equation:

$2 M n O_{4}^{2 -} + 10 H^{+} + 6 e^{-} \to M n_{2} O_{3} + 5 H_{2} O$ $2MnO_4^(2-) + 10H^(+) + 6e^(-) rarr Mn_2O_3 +5H_2O$ $(i)$ $(i)$

Oxidation half equation:

$H O (O =) C - C (= O) O H \to 2 C O_{2} ↑ ⏐ + 2 H^{+} + 2 e^{-}$ $HO(O=)C-C(=O)OH rarr 2CO_2uarr +2H^(+) +2e^(-)$ $(i i)$ $(ii)$

So $(i) + 3 \times (i i) =$ $(i)+3xx(ii)=$

$2 M n O_{4}^{2 -} + 3 {H O (O =) C}_{2} + 4 H^{+} \to M n_{2} O_{3} + 5 H_{2} O + 6 C O_{2}$ $2MnO_4^(2-) + 3{HO(O=)C}_2 +4H^(+)rarr Mn_2O_3 +5H_2O +6CO_2$

Which (I think) is balanced with respect to mass and charge as required.

You've got potassium oxide in your equation, for which I have not accounted. The redox couple is manganese and carbon.

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