how to balance the following by oxidation number method MnO4·(aqJ + SO2(g) ._ Mn2+(aq) + HSO4·

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Solution

First split the question up to look like this.

SO2 ---> (SO4)2-

MnO4- ---> (Mn)2+

You don't need to balance for S or for Mn so start with oxygen on each side. Then balance for hydrogen on each equation. Finally, put both together so your total charges cancel out (system of equations sort of).

After that it's just simplification.

SO2 + 2H2O ---> SO4(2-) + 4H+ +2e- ] Multiply by factor of 5

MnO4- + 8H+ +5e----> Mn(2+) + 4H2O ] Multiply by factor of 2

5SO2 + 10H2O ----> 5SO4 + 20H + 10e-
2MnO4 + 16H + 10e- ----> 2Mn(2+) +8H2O

electrons cancel out, now just simplify it all out.

5SO2 + 2MnO4(-) + 2H2O ----> 5SO4(-) + 2Mn(2+) + 4H2O

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Similar questions

Balance the following redox reactions by ion-electron method:

(a) (aq) + I^– (aq) → MnO₂ (s) + I₂(s) (in basic medium)

(b) (aq) + SO₂ (g) → Mn²⁺ (aq) + (aq) (in acidic solution)

(d) + SO₂(g) → Cr³⁺ (aq) + (aq) (in acidic solution)

M n O_{4}^{-} (a q) + C_{2} H_{2} O_{4} (a q) \to M n^{2 +} (a q) + C O_{2} (g) + H_{2} O (l)

B r^{-} + M n O_{4}^{-} \to B r O_{3}^{-} + M n O_{2}

M n O_{4}^{-} (a q) + S O_{2} (g) \to M n^{2 +} (a q) + H S O_{4}^{-} (a q)

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

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