8.1: Balancing redox reactions
- Page ID
- 164370
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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 begin by learning to balance 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 foolproof, systematic method. The electron-ion method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples.
Example 1 - Acidic Solution
I- is oxidized to IO3- by MnO4-, which is reduced to Mn2+. This reaction takes place in acidic solution.
How can this reaction be balanced? In the electron-ion method we follow a series of ten steps:
Step 1: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.
- MnO4- + I- → IO3- + Mn2+ (acid)
Step 2: Assign the oxidation states to each element on both sides of the equation.
- MnO4- Mn = +7 O = -2
- I- I = -1
- IO3- I = +5 O = -2
- Mn2+ Mn = +2
Step 3: Separate the reaction into two half reactions, balancing the element undergoing redox in each.
red. MnO4- → Mn2+ Manganese is being reduced
ox. I- → IO3- Iodine is being oxidized
Step 4: Add the appropriate number of electrons to each half-reaction. Electrons are always reactants in the reduction half-reaction and products in the oxidation half reaction.
red. MnO4- + 5 e- → Mn2+ Mn is reduced from +7 to +2. This adds 5 electrons.
ox. I- → IO3- + 6 e- I is oxidized from -1 to +5. This removes 6 electrons.
Step 5: Balance the charge of each half-reaction by adding hydrogen ions, H+. Since this reaction occurs in acidic solution, there are ample H+ ions for charge balance.
red. MnO4- + 5 e- + 8H+ → Mn2+ Both sides of the equation have a total 2+ charge.
ox. I- → IO3- + 6 e- + 6 H+ Both sides of the equation have a total 1- charge.
Step 6: Balance the oxygen and hydrogen atoms by adding H2O to the appropriate side of each half-reaction.
red. MnO4- + 5 e- + 8 H+ → Mn2+ + 4H2O
ox. I- + 3 H2O → IO3- + 6 e- + 6 H+
Step 7: Multiply each half-reaction by the appropriate coefficient so that there are equal numbers of electrons as reactants and products.
red. (MnO4- + 5 e- + 8 H+ → Mn2+ + 4H2O) x 6
ox. (I- + 3 H2O → IO3- + 6 e- + 6 H+) x 5
Step 8: Combine the half-reactions into an overall equation.
6 MnO4- + 5 I- + 30 e- + 48 H+ + 15 H2O → 6 Mn2+ + 5 IO3- + 30 e- + 30 H+ + 24 H2O
Step 9: Cancel the H+, electrons, and H2O:
6 MnO4- + 5 I- + 30 e- + 48 H+ + 15 H2O → 6 Mn2+ + 5 IO3- + 30 e- + 30 H+ + 24 H2O
The overall balanced reaction is therefore:
6 MnO4-(aq) + 5 I-(aq) + 18 H+(aq) → 6 Mn2+(aq) + 5 IO3-(aq) + 9 H2O(ℓ)
Step 10: Check your work by making sure that all elements and charges are balanced.
Example 2 - Basic Solution
The chromate ion (CrO42-) oxidizes Fe(OH)2 to Fe(OH)3. The CrO42- ion is reduced to Cr(OH)3. This reaction takes place in basic solution.
How can this reaction be balanced? In the electron-ion method we follow a series of ten steps:
Step 1: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.
- CrO42- + Fe(OH)2 → Cr(OH)3 + Fe(OH)3 (basic)
Step 2: Assign the oxidation states to each element on both sides of the equation.
- CrO42- Cr = +6 O = -2
- Fe(OH)2 Fe = +2 O = -2 H = +1
- Cr(OH)3 Cr = +3 O = -2 H = +1
- Fe(OH)3 Fe = +3 O = -2 H = +1
Step 3: Separate the reaction into two half reactions, balancing the element undergoing redox in each.
red. Cr O4- → Cr(OH)3 Chromium is being reduced
ox. Fe(OH)2 → Fe(OH)3 Iron is being oxidized
Step 4: Add the appropriate number of electrons to each half-reaction. Electrons are always reactants in the reduction half-reaction and products in the oxidation half reaction.
red. CrO42- + 3 e- → Cr(OH)3 Cr is reduced from +6 to +3. This adds 3 electrons.
ox. Fe(OH)2 → Fe(OH)3 + e- Fe is oxidized from +2 to +3. This removes 1 electron.
Step 5: Balance the charge of each half-reaction by adding hydroxide ions, OH-. Since this reaction occurs in basic solution, there are ample OH- ions for charge balance.
red. CrO42- + 3 e- → Cr(OH)3 + 5 OH- Both sides of the equation have a total 5- charge.
ox. Fe(OH)2 + OH- → Fe(OH)3 + e- Both sides of the equation have a total 1- charge.
Step 6: Balance the oxygen and hydrogen atoms by adding H2O to the appropriate side of each half-reaction.
red. CrO42- + 3 e- + 4 H2O → Cr(OH)3 + 5 OH-
ox. Fe(OH)2 + OH- → Fe(OH)3 + e- This half-reaction is already balanced.
Step 7: Multiply each half-reaction by the appropriate coefficient so that there are equal numbers of electrons as reactants and products.
red. CrO42- + 3 e- + 4 H2O → Cr(OH)3 + 5 OH-
ox. (Fe(OH)2 + OH- → Fe(OH)3 + e-) x 3
Step 8: Combine the half-reactions into an overall equation.
CrO42- + 3 Fe(OH)2 + 3 e- + 3 OH- + 4 H2O → Cr(OH)3 + 3 Fe(OH)3 + 3 e- + 5 OH-
Step 9: Cancel the OH-, electrons, and H2O:
CrO42- + 3 Fe(OH)2 + 3 e- + 3 OH- + 4 H2O → Cr(OH)3 + 3 Fe(OH)3 + 3 e- + 5 OH-
The overall balanced reaction is therefore:
CrO42-(aq) + 3 Fe(OH)2(s) + 4 H2O(ℓ) → Cr(OH)3(s) + 3 Fe(OH)3(s) + 2 OH-(aq)
Step 10: Check your work by making sure that all elements and charges are balanced.
Contributors
Adapted from the Wikibook constructed by Chemistry 310 students at Penn State University.