8.1: Balancing redox reactions

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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 (CrO₄^2-) oxidizes Fe(OH)₂ to Fe(OH)₃. The CrO₄^2- 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.

CrO₄^2- + 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.

CrO₄^2- + 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:

CrO₄^2- + 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:

CrO₄^2-(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.