Question
Balance the following ionic equations. $\text{MnO}^{-}_4+\text{SO}^{2-}_3+\text{H}^+\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{Mn}^{2+}+\text{SO}^{2-}_4+\text{H}_2\text{O}$
✓
Answer
Dividing the equation into two half reactions:
Oxidation half reaction: $\text{SO}^{2-}_3\xrightarrow{ \ \ \ \ \ \ \ }\text{SO}^{2-}_4$
Reduction half reaction: $\text{MnO}^-_4\xrightarrow{ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
Balancing oxidation and reduction half reactions separately as:
Oxidation half reaction
$\text{SO}^{2-}_3\xrightarrow{ \ \ \ \ \ \ \ }\text{SO}^{2-}_4$
$\text{SO}^{2-}_3\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{SO}^{2-}_4+2\text{e}^{-}$
Since the reaction occurs in acidic medium,
$\text{SO}^{2-}_3\xrightarrow{ \ \ \ \ \ \ \ \ }\text{SO}^{2-}_4+2\text{e}^-+2\text{H}^+$
$\text{SO}^{2-}_3+\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ }\text{SO}^{2-}_4+2\text{H}^++2\text{e}^-...\text{(i)}$
Reduction half reaction
$\text{MnO}^-_4\xrightarrow{ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
$\text{MnO}^-_4+5\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
$\text{MnO}^-_4+8\text{H}^++5\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O}...(\text{ii})$
To balance the electrons, multiply eq. (i) by 5 and eq. (ii) by 2 and add
$2\text{MnO}^-_4+5\text{SO}^{2-}_3+6\text{H}^+\xrightarrow{ \ \ \ \ \ \ \ }2\text{Mn}^{2+}5\text{SO}^{2-}_4+3\text{H}_2\text{O}$
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