For the unbalanced reaction:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{Mn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s}) \]
- In $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$: Cr has oxidation state +6
- In $\mathrm{Mn}^{2+}$: Mn has oxidation state +2
- In $\mathrm{Cr}^{3+}$: Cr has oxidation state +3
- In $\mathrm{MnO}_{2}$: Mn has oxidation state +4
Reduction half-reaction:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+} \]
Oxidation half-reaction:
\[ \mathrm{Mn}^{2+} \rightarrow \mathrm{MnO}_{2} \]
For Cr reduction:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+} \]
Balance oxygen by adding H₂O:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \]
Balance hydrogen by adding H⁺:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \]
Balance charge by adding electrons:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \]
For Mn oxidation:
\[ \mathrm{Mn}^{2+} \rightarrow \mathrm{MnO}_{2} \]
Balance oxygen by adding H₂O:
\[ \mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{2} \]
Balance hydrogen by adding H⁺:
\[ \mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{H}^{+} \]
Balance charge by adding electrons:
\[ \mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{H}^{+} + 2e^{-} \]
To balance electrons, multiply the oxidation half-reaction by 3:
\[ 3(\mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{H}^{+} + 2e^{-}) \]
\[ 3\mathrm{Mn}^{2+} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 3\mathrm{MnO}_{2} + 12\mathrm{H}^{+} + 6e^{-} \]
Now combine with the reduction half-reaction:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} + 3\mathrm{Mn}^{2+} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 3\mathrm{MnO}_{2} + 12\mathrm{H}^{+} + 6e^{-} \]
Cancel out 6e⁻ from both sides:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 3\mathrm{Mn}^{2+} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 3\mathrm{MnO}_{2} + 12\mathrm{H}^{+} \]
Cancel out 12H⁺ from both sides:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 2\mathrm{H}^{+} + 3\mathrm{Mn}^{2+} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 3\mathrm{MnO}_{2} \]
Cancel out 6H₂O from both sides:
\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 2\mathrm{H}^{+} + 3\mathrm{Mn}^{2+} \rightarrow 2\mathrm{Cr}^{3+} + \mathrm{H}_{2}\mathrm{O} + 3\mathrm{MnO}_{2} \]
In the balanced equation, the coefficient of water is 1.
The coefficient for water when the reaction is balanced in acidic solution is \( \boxed{1} \).
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