05. Balancing Redox Reactions using Half-Reaction Method

Please find below the instructions on how to balance acidic (H⁺) and basic (OH^-) reduction-oxidation reactions. Most redox reactions can't be balanced by inspection and therefore, require a set of steps to account for the movement of electrons taking place during the reaction. Unless otherwise stated, it is assumed that all redox reactions are in an aqueous acidic environment. Therefore, the addition of water and H⁺ to the reaction to help balance it are allowed. Please see below for one of the many methods available for balancing a redox reaction. It's known as the half-reaction method.

The reaction used below is from the Periodic Properties lab.

How to balance an acidic redox reaction using the half-reaction method:

Reaction to be balanced:

MnO₄^-1_(aq) + I^-_(aq) → MnO_2(s) + I_2(aq)

(Notice this reaction has no spectator ions; potassium (K⁺) and sodium (Na⁺) have been removed.)

• 1. First identify the species being oxidized and reduced in the reaction. You can do this by identifying the oxidation and reduction state of the atoms involved, such as the oxidation state of oxygen in a compound is typically -2 and since the oxidation states must add up to the overall charge of the species, the oxidation state of Mn in MnO₄^-1 is +7.

  2. The oxidation states of I and Mn in I^-, MnO₂, and I₂ are -1, +4, and 0, respectively.

  3. Next, determine which species are being oxidized and which are being reduced based on the oxidation states for all atoms. Then, separate the equation into two half reactions--an oxidation half-reaction and a reduction half-reaction.

     1. Oxidation half reaction: I^-_(aq) →I_2(aq)

     2. Reduction half reaction: MnO₄^-1_(aq) →MnO_2(s)

1. Balance the atoms going through oxidation and reduction, in this case--Mn and I.

   1. Oxidation half reaction: 2 I^-_(aq) →1 I_2(aq)

      1. Reduction half reaction: 1 MnO₄^-1_(aq) → 1 MnO_2(s)

2. If there are unbalanced oxygen atoms in your half reactions, add H₂O_(l) to balance the oxygen atoms.

• 1. Oxidation half reaction: 2 I^-_(aq) →1 I_2(aq)

     1. Reduction half reaction: 1 MnO₄^-1_(aq) → 1 MnO_2(s)+ 2H₂O_(l)

3. By adding water, you may have unbalanced the number of protons in the reaction, so now add H⁺_(aq) to balance hydrogens. (Note: it's assumed this is an acidic reaction due to no mention of the acidity or basicity of the reaction.)

• • 1. Oxidation half reaction: 2 I^-_(aq) →1 I_2(aq)

    2. Reduction half reaction: 8H⁺_(aq) + 1 MnO₄^-1_(aq) → 1 MnO_2(s)+ 2H₂O_(l)

4. Then, balance the charge in each half-reaction using electrons (e^-). The charge does not have to be zero on both sides of the reaction...it just needs to be the same.

• 1. Oxidation half reaction: 2 I^-_(aq) →1 I_2(aq)+ 2e^-

     1. Reduction half reaction: 3e^- + 4H⁺_(aq) + 1 MnO₄^-1_(aq) → 1 MnO_2(s)+ 2H₂O_(l)

5. You now have two balanced half-reactions. YET, before the half reactions can be added back together to create an overall balanced redox reaction, you may need to modify each half-reaction to ensure that the overall reaction will consume the same number of electrons that are being produced. In this case, we have two electrons produced in the oxidation half reaction and three electrons consumed in the reduction half reaction. The least common multiple is 6. So, we need to multiply each half-reactant by appropriate coefficients so that 6 electrons are exchanged.

• 1. Oxidation half reaction: [ 2 I^-_(aq) →1 I_2(aq)+ 2e^-] x 3

     1. 6I^-_(aq)→ 3I_2(aq) + 6e^-

     2. Reduction half reaction: [3e^- + 4H⁺_(aq) + 1 MnO₄^-1_(aq) → 1 MnO_2(s)+ 2H₂O_(l)] x 2

     3. 6e^- + 8H⁺_(aq) + 2MnO₄^-1_(aq) → 2MnO_2(s)+ 4H₂O_(l)

  2. Finally, combine the two half reactions to give the overall balanced acidic redox reaction. Consolidate or cancel out any species, such as the electrons, water, or protons, that are on both sides of the reaction. Verify that the numbers of atoms and the charges are balanced.

Balanced redox reaction: 8H⁺_(aq) + 6I^-_(aq) + 2MnO₄^-1_(aq) → 3I_2(aq) + 2MnO_2(s)+ 4H₂O_(l)

How to balance a BASIC redox reaction using the half-reaction method:

If the reaction given is stated to be in a basic solution (excess OH^-), then perform the following conversion of any balanced acidic redox reaction, such as the one above:

Use one of the following reactions to help you cancel out the protons (H⁺) in your balanced acidic reaction and leave the reaction with excess hydroxide ions (OH^-.)

For the reaction in step 8 above, we need to cancel out 8H⁺ ions that appear on the reactant side of the equation. Therefore, we will need the reaction on the right to occur 8 times when adding it to our overall balanced reaction to cancel out all the protons.

The 8 H⁺ ions cancel out and you can consolidate the 8H₂O molecules with the 4H₂O molecules to generate the final balanced BASIC redox reaction:

Lastly, please always verify that your final reaction is balanced by checking if the charge (or any specific atom) is balanced on both sides of the reaction.

Here are some websites with directions and practice problems for balancing both acidic and basic redox reactions:

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions

http://pages.towson.edu/ladon/redoxa.html

http://www.sciencegeek.net/APchemistry/APpdfs/extraredox.pdf

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/oxred_2.php

See below for an image of 10 easy steps to balancing redox reaction from lab!

See below for a summary worksheet on net ionic equations, redox reactions, and uncertainty analysis.