6.3.2: Balancing Redox Reactions
- Page ID
- 301382
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In studying redox chemistry, it is important to remember balancing electrochemical reactions. Simple redox reactions (for example, H2 + I2 → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a systematic method. There are many different methods for doing this. The focus in this section will be on the ion-electron method allows one to balance redox reactions regardless of their complexity and allows for identification of the species changing valence states in the reaction. We illustrate this method with two examples.
Example 1:
I- is oxidized to IO3- by MnO4-, which is reduced to Mn2+.
How can this reaction be balanced? In the ion-electron method we follow a series of four steps:
Step 1A: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.
- MnO4- + I- → IO3- + Mn2+ (The elements undergoing redox are Mn and I)
Step 1B: Separate the reaction into two half reactions, balancing the element undergoing redox in each.
- MnO4- → Mn2+
- I- → IO3-
Step 2A: Balance the oxygen atoms by adding water to one side of each half reaction.
- MnO4-→ Mn2+ + 4H2O
- 3H2O + I- → IO3-
Step 2B: Balance the hydrogen atoms by adding H+ ions.
- 8H+ + MnO4-→ Mn2+ + 4H2O
The left side has a net charge of +7 and the right side has a net charge of +2
- 3H2O + I- → IO3- + 6H+
The left side has a net charge of -1 and the right side has a net charge of +5
Step 2C: Balance the overall charge by adding electrons
- 8H+ + 5e- + MnO4-→ Mn2+ + 4H2O
The left side has a charge of +2 while the right side has a charge of +2. They are balanced.
- 3H2O + I- → IO3- + 6H+ + 6e-
The left side has a charge of -1 while the right side has a charge of -1. They are balanced.
Note: We did not need to explicitly determine the oxidation states of Mn or I to arrive at the correct number of electrons in each half reaction.
Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides
- 6 (8H+ + 5e- + MnO4-→ Mn2+ + 4H2O)
- 5 (3H2O + I- → IO3- + 6H+ + 6e-)
48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+
Cancel the H+, electrons, and water:
48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+
The overall balanced reaction is therefore:
18H+ + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O
Check your work by making sure that all elements and charges are balanced.
Step 4: If the reaction occurs under basic conditions, we add OH- to each side to cancel H+
18H+ + 18OH- + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O + 18OH-
The 18H+ + 18OH- will become 18H2O so the overall balanced reaction is:
9H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 18OH-
Again, it is a good idea to check and make sure that all of the elements are balanced, and that the charge is the same on both sides. If this is not the case, you need to find the error in one of the earlier steps.
Example 2:
Redox reaction of S2O32- and H2O2
S2O32- + H2O2 → S4O62- + H2O
Which elements are undergoing redox? S and O
Step 1: Write out half reactions, balancing the element undergoing redox
- 2S2O32- → S4O62-
- H2O2 → 2H2O
Step 2A: Balance oxygen (already balanced)
Step 2B: Balance hydrogen:
- 2S2O32- → S4O62-
- H2O2 + 2H+ → 2H2O
Step 2C: Balance charge by adding electrons:
- 2S2O32- → S4O62- + 2e-
- H2O2 + 2H+ + 2e- → 2H2O
Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides (already equal)
Overall balanced reaction:
2S2O32- + H2O2 + 2H+ → S4O62- + 2H2O
Note that again, we did not need to know the valence states of S or O in the reactants and products in order to balance the reaction. In this case, assigning the valence states is a bit complex, because S4O62- contains sulfur in more than one valence state.