Question 16.8: Balancing Redox Reactions Occurring in Basic Solution Balanc......
Balancing Redox Reactions Occurring in Basic Solution
Balance the redox reaction occurring in basic solution.
\mathrm{CN}^{-}(a q)\,+\,\mathrm{MnO}_{4}^{-}(a q)\,\longrightarrow\,\mathrm{CNO}^{-}(a q)\,+\,\mathrm{MnO}_{2}(s) (basic solution)
1. Follow the half-reaction method for balancing redox reactions. Begin by assigning oxidation states.
2. Separate the overall reaction into two half-reactions.
OXIDATION {\mathrm{CN}}^{-}(a q)\longrightarrow{\mathrm{CNO}}^{-}(a q)
REDUCTION \mathrm{MnO_{4}}^{-}\left(aq\right)\longrightarrow\mathrm{MnO_{2}}(s)
3. Balance each half-reaction with respect to mass.
• Balance all elements other than H and O.
• Balance O by adding \mathrm{H}_{2}\mathrm{O}.
• Balance H by adding \mathrm{H}^{+}.
• Neutralize \mathrm{H}^{+} by adding \mathrm{OH}^{-}. Add the same number of \mathrm{OH}^{-} to each side of the equation (to preserve mass balance).
• Cancel any water molecules that occur on both sides of the half-reaction.
All elements other than H and O are already balanced.
\mathrm{CN}^{-}(a q)+\Pi_{2}\mathrm{O}(l)\longrightarrow\mathrm{CNO}^{-}(a q)\mathrm{MnO_{4}}^{-}\left(a q\right)\longrightarrow\mathrm{MnO_{2}}(s)+2\,\mathrm{H_{2}}O(l)
\mathrm{CN}^{-}\left(a q\right)+\mathrm{H}_{2}O(l)\longrightarrow\mathrm{CNO}^{-}\left(a q\right)+2~\mathrm{H}^{+}\left(a q\right)
\mathrm{MnO_{4}}^{-}(a q)+4\,\mathrm{H^{+}}(a q)\,\longrightarrow\,\mathrm{MnO_{2}}(s)+2\,\mathrm{H_{2}}\mathrm{O}(l)
\mathrm{CN}^{-}(a q)+\mathrm{H}_{2}{O}(l)+2\,{\mathrm{OH}^{-}}\,(a q)\longrightarrow\mathrm{CNO}^{-}(a q)+\underbrace{2\,\mathrm{H}^{+}(a q)+2\,\mathrm{OH^{-}}\,(a q)}_{2\,{\mathrm{H}}_{2}{\mathrm{O}}(l)}
\mathrm{MnO_{4}}^{-}(a q)\,+\,\underbrace{4\,\mathrm{H}^{+}(a q)\,+\,4\,\mathrm{OH}^{-}(a q)}_{{4\,\mathrm{H}_{2}\mathrm{O}(l)}}\longrightarrow\mathrm{MnO_{2}(s)\,+\,2\,H_{2}O(l)\,+\,4\,O H}^{-}(a q)
\mathrm{CN}^{-}(a q)+\cancel{\mathrm{H}_{2}\mathrm{O}}(l)+2\,\mathrm{OH}^{-}(a q)\,\longrightarrow\,\mathrm{CNO}^{-}(a q)+\cancel{2}\,\mathrm{H}_{2}\mathrm{O}(l)
\mathrm{MnO_{4}}^{-}(a q)+2\,\mathrm{\cancel{4}\,H_{2}O(}l)\,\longrightarrow\,{\mathrm{MnO_{2}}}(s)\,+\,\cancel{2\,\mathrm{H}_{2}O}(l)\,+\,4\,\mathrm{OH}^{-}(a q)
4. Balance each half-reaction with respect to charge.
\mathrm{CN}^{-}(a q)\,+\,2\,\mathrm{OH}^{-}(a q)\,\longrightarrow\,\mathrm{CNO}^{-}(a q)\,+\,\mathrm{H}_{2}O(l)\,+\,2\mathrm{e^{-}}3~\mathrm{e}^{-}\,+\,\mathrm{MnO}_{4}^{-}(a q)\;+\,2\,\mathrm{H}_{2}\mathrm{O}(l)\,\longrightarrow\,\mathrm{MnO}_{2}(s)\,+\,4\,\mathrm{OH}^{-}(a q)
5. Make the number of electrons in both half-reactions equal.
3\times\mathrm{[CN^{-}(a q)\,+\,2\,O H^{-}(a q)\,\longrightarrow\,C N O^{-}(a q)\,+\,H_{2}O(l)\,+\,2e^{-}]}2\,\times\,[3{\mathrm{e}}^{-}\,+\,\mathrm{MnO_{4}}^{-}(a q)\,+\,2\,\mathrm{H}_{2}O(l)\,\longrightarrow\,\mathrm{MnO_{2}}(s)\,+\,4\,\mathrm{OH}^{-}(a q)]
6. Add the half-reactions together and cancel.
3~\mathrm{CN}^{-}(a q)+\cancel{\,6~{\mathrm{OH}}}\left(a q\right)\longrightarrow3~\mathrm{CNO}^{-}(a q)+\cancel{3~\mathrm{H}_{2}\mathrm{O}}(l)+\,\cancel{6}\mathrm{e}^{-}\underline{{{\mathrm{\cancel{6}~e}}}^{-}\;+\;2\;\mathrm{MnO}_{4}^{-}(a q)\,+\,\cancel{4}\;\mathrm{H}_{2}O(l)\;\longrightarrow\;2\;\mathrm{MnO}_{2}(s)\;+\;2\;\cancel{8}\;\mathrm{OH}^{-}(a q)}
3~\mathrm{CN}^{-}(a q)+2~\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\longrightarrow3~\mathrm{CNO}^{-}(a q)+2~\mathrm{MnO}_{2}(s)+2~\mathrm{OH}^{-}(a q)
7. Verify that the reaction is balanced.
| Reactants | Products |
| 3 C | 3 C |
| 3 N | 3 N |
| 2 Mn | 2 Mn |
| 9 O | 9 O |
| 2 H | 2 H |
| -5 charge | -5 charge |