Question 16.6: Balancing Redox Equations Using the Half-Reaction Method Bal......
Balancing Redox Equations Using the Half-Reaction Method
Balance the redox reaction.
1. Assign oxidation states to all atoms and identify the substances being oxidized and reduced.
2. Separate the overall reaction into two half-reactions, one for oxidation and one for reduction.
OXIDATION \mathrm{F}e^{2+}(a q)\longrightarrow\mathrm{Fe}^{3+}(a q)
REDUCTION
{\mathrm{MnO}}_{4}^{~-}(a q)\longrightarrow\mathrm{Mn}^{2+}(aq)
3. Balance each half-reaction with respect to mass in the following order:
• Balance all elements other than H and O.
• Balance O by adding \mathrm{H}_{2}\mathrm{O}.
• Balance H by adding \mathrm{H}^{+}.
All elements other than hydrogen and oxygen are balanced, so you can proceed to the next step.
\mathrm{Fe}^{2+}(a q)\longrightarrow\mathrm{Fe}^{3+}(a q)
{\mathrm{MnO_{4}}}^{-}(a q)\longrightarrow
\mathrm{Mn}^{2+}(a q)+\mathrm{4~H}_{2}\mathrm{O}(l)
\mathrm{Fe}^{2+}(a q)\longrightarrow\mathrm{Fe}^{3+}(a q)
{\mathrm{8~H}}^{+}(a q)+{\mathrm{M n O_{4}}}^{-}(a q)\longrightarrow
\mathrm{Mn}^{2+}(a q)+4\,\mathrm{H_{2}O}(l)
4. Balance each half-reaction with respect to charge by adding electrons to the right side of the oxidation half-reaction and the left side of the reduction half-reaction. (The sum of the charges on both sides of each equation should be equal.)
\mathrm{Fe}^{2+}(a q)\longrightarrow\mathrm{Fe}^{3+}(a q)+ 1~\mathrm{e}^{-}5\mathrm{e}^{-}+8\;\mathrm{H}^{+}(a q)+\mathrm{MnO_{4}}^{-}(a q)\;\longrightarrow
\mathrm{Mn}^{2+}(a q)~+~4~\mathrm{H}_{2}\mathrm{O}(l)
5. Make the number of electrons in both half-reactions equal by multiplying one or both halfreactions by a small whole number.
5~\times~[\mathrm{Fe}^{2+}(a q){\longrightarrow}\mathrm{Fe}^{3+}(a q)+1~\mathrm{e}^{-}]5\mathrm{e}^{-}+8\;\mathrm{H}^{+}(a q)+\mathrm{MnO_{4}}^{-}(a q)\;\longrightarrow
\mathrm{Mn}^{2+}(a q)~+~4~\mathrm{H}_{2}\mathrm{O}(l)
6. Add the two half-reactions together, canceling electrons and other species as necessary.
5\,{\mathrm{F e}}^{2+}(a q)\,\longrightarrow\,5\,{\mathrm{F e}}^{3+}(a q)+\cancel{5\mathrm{e^{-}}}
\cancel{5~\mathrm{e}^{-}}+{\mathrm{8~H}}^{+}(a q)+{\mathrm{M n O_{4}}}^{-}(a q)\longrightarrow
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathrm{Mn}^{2+}(a q)~+4~\mathrm{H}_{2}\mathrm{O}(l)}
5\;\mathrm{Fe}^{2+}(a q)+8\;\mathrm{H}^{+}(a q)+\mathrm{MnO_{4}}^{-}(a q)
\longrightarrow5\,\mathrm{Fe}^{3\,+}(a q)+\mathrm{Mr}^{2+}(a q)+\mathrm{~4~H_{2}O}(l)
7. Verify that the reaction is balanced with respect to both mass and charge.
Reactants Products
5 Fe 5 Fe
8 H 8 H
1 Mn 1 Mn
4 O 4 O
+17 charge +17 charge