Balance the following redox reactions by ion electron method- Cr2O2-7-SO2g- Cr3-aq-SO2-4aq in acidic solution

The oxidation half reaction is SO_2(g) + 2H_2O(l) → SO_4^2 + 4H^+(aq) +2e^ .The reduction half reaction is Cr_2O_7^2 (aq) + 14H^+(aq) +6e^ → 2Cr^3+ ...

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Standard XII

Chemistry

Stoichiometric Calculations

Balance the f...

Question

Balance the following redox reactions by ion electron method.
$C r_{2} O_{7}^{2 -} + S O_{2} (g) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)$ (in acidic solution)

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Solution

The oxidation half reaction is $S O_{2} (g) + 2 H_{2} O (l) \to S O_{4}^{2 -} + 4 H^{+} (a q) + 2 e^{-}$ .
The reduction half reaction is $C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$ .
The oxidation half reaction is multiplied by 3 and added to the reduction half reaction to obtain the balanced redox reaction.
$C r_{2} O_{7}^{2 -} + 3 S O_{2} (g) + 2 H^{+} (a q) \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + H_{2} O (l)$

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Balance the following redox reactions by ion electron method.Cr2O2−7+SO2(g)→Cr3+(aq)+SO2−4(aq) (in acidic solution)

Balance the following redox reactions by ion electron method.
$C r_{2} O_{7}^{2 -} + S O_{2} (g) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)$ (in acidic solution)