Question
Balance the following redox reactions by ion-electron method:
$\text{Cr}_2\text{O}_7^{2-}+\text{SO}_2(\text{g})\rightarrow\text{Cr}^{3+}(\text{aq})+\text{SO}^{2-}_4(\text{aq})$ (in acidic solution).

Answer

Following the steps as in part (a), we have the oxidation half reaction as:
$\text{SO}_{2(\text{g})}+2\text{H}_2\text{O}_{(\text{l})}\rightarrow\text{SO}^{2-}_{4(\text{aq})}+4\text{H}^+_{(\text{aq})}+2\text{e}^-$
And the reduction half reaction as:
$\text{Cr}_2\text{O}^{2-}_{7(\text{aq})}+14\text{H}^{+}_{(\text{aq})}+6\text{e}^-\rightarrow2\text{Cr}^{3+}_{(\text{aq})}+7\text{H}_2\text{O}_{(\text{l})}$
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
$\text{Cr}_2\text{O}^{2-}_{7(\text{aq})}+3\text{SO}_{2(\text{g})}+2\text{H}^{+}_{(\text{aq})}\rightarrow2\text{Cr}^{3+}_{(\text{aq})}+3\text{SO}^{2-}_{4(\text{aq})}+\text{H}_2\text{O}_{(\text{l})}$

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