Question
Balance the following ionic equations.
$\text{MnO}^-_4+\text{H}^++\text{Br}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Mn}^{2+}+\text{Br}_2+\text{H}_2\text{O}$

Answer


Dividing the equation into two half reactions:
Oxidation half reaction: $\text{Br}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Br}_2$
Reduction half reaction: $\text{MnO}^-_4\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
Balancing oxidation and reduction half reactions separately as:
Oxidation half reaction
$\text{Br}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Br}_2$
$2\text{Br}^-\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{Br}_2$
$2\text{Br}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Br}_2+2\text{e}^-...(\text{i})$
Reduction half reaction
$\text{MnO}^-_4\xrightarrow{ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
$\text{MnO}^-_4+5\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
$\text{MnO}^-_4+8\text{H}^++5\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{Mn}^{2+}$
$\text{MnO}^-_4+8\text{H}^++5\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O}...(\text{ii})$
To balance the electrons, multiply eq. (i) by 5 and eq. (ii) by 2 and add
$2\text{MnO}^-_4+10\text{Br}^-+16\text{H}^+\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{Mn}^{2+}+5\text{Br}_2+8\text{H}_2\text{O}$

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