$M n O_{4}^{⊖} + S O_{3}^{2 -} ⟶ S O_{4}^{2 -} + M n O_{2}$

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Solution

The balanced redox reaction is as follows:
$2 H^{\oplus} + 2 M n O_{4}^{⊖} + 3 S O_{3}^{2 -} ⟶ 2 M n O_{2} + 3 S O_{4}^{2 -} + H_{2} O$

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Similar questions

C r_{2} O_{7}^{2 -} + H^{+} + I^{-} \to C r^{3 +} + I_{2} + H_{2} O

C r_{2} O_{7}^{2 -} + F e_{2}^{+} + H^{+} \to C r^{3 +} + F e^{3 +} + H_{2} O

M n O^{4 -} + S O_{3}^{2 -} + H^{+} \to M n^{2 +} + S O_{4}^{2 -} + H_{2} O

M n O^{4 -} + H^{+} + B r^{-} \to M n^{2 +} + B r_{2} + H_{2} O

C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to C r^{+ 3} + S O_{4}^{2 -}

C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to S O_{4}^{2 -} + C r^{3 +}

0.1 M K M n O_{4}

m L

0.158 g

N a_{2} S_{2} O_{3}

{S_{2} O_{3}}^{2 -} + {M n O_{4}}^{-} + H_{2} O \to M n O_{2} (s) + {S O_{4}}^{2 -} + O H^{-}

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

E^{\circ} = 1.51 V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O

E^{\circ} = 1.23 V

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