3. Balancing redox equations• We have seen in Chemistry 11 that to balance a chemical equation,

we would have to consider the following four criteria:1. Mass2. Electric charges3. Number of atoms4. Energy

• These criteria arise from the conservation laws of mass, energy and charge.

• They can be applied to most of the chemical equations.

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• But for balancing redox equations, in addition to these four conditions, the conservation of number of transferred electronsshould also be considered.

• For example:

• Note that Zn is a reducing agent giving up 2 electrons, while Cu+ is an oxidizing agent taking up 1 electron. It implies that the mole ratio of Zn to Cu+ is 1:2.

• Hence,

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• Some redox equations can be balanced simply by inspection. For more complicated ones, we can make use of half-reactions.

• For the previous example, we split the reaction into the corresponding oxidation and reduction half-reactions:

• The final, balanced equation in obtained by balancing the electrons transferred.

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• To balance a half reaction, we follow the general steps below.

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1. Balance all major atoms other than oxygen and hydrogen.

2. Balance oxygens by adding watermolecules if necessary.

3. Assume that solutions are acidic and balance hydrogens by adding H+ ions.

4. Balance the charge by adding necessary electrons.

• Example: Balance .

• Solution:• Step 1 – balance Cr:

• Step 2 – balance O:

• Step 3 – balance H:

• Step 4 – balance electrons:

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• Example: Balance in basic solution.

• Solution:• Step 1 – balance Mn:

• Step 2 – balance O:

• Step 3 – balance H:

• Step 4 – balance electrons:

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• Since the redox reaction occurs in a basic solution, H+ ions on the left hand side have to be removed by adding OH- ions.

• Step 5 – add :

• Step 6 – balance water:

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• Example: Balance the following redox reaction:

• Solution: We split it into two half reactions and balance them one by one.

• Oxidation:

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• Reduction:

• Now we combine the two balanced half reactions together and balance the electrons transferred:

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• Practice: Balance the following redox reaction in acidic solution.

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• In basic solutions, the final equation can be converted by adding enough number of hydroxide ions to remove H+ ions.

• In the previous example, we have obtained the following equation in acidic solution:

• To remove H+, we add 8 hydroxide ions on both sides:

• Then we cancel out water to yield the final equation:

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• Practice: Balance the following redox reaction in basic solution.

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• In certain redox reactions, one species may undergo oxidation and reduction simultaneously. Such a reaction is called disproportionation.

• For example:

• To balance this reaction, we will have to consider the following half reactions:

• After balancing we obtain the final equation

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• Practice: Balance the following redox reaction in base:

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• An alternative method, which is conceptually more appealing, of balancing redox equations relies on the change in oxidation number.

• For example, consider the following reaction:

• Note that the change of ON indicates how many electrons are transferred to/from a species, and this quantity is conserved in a balanced redox reaction.

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+7 +2+4 +5

• To equalize the electrons transferred, we multiply appropriate coefficients to the species in the equation:

• Then we balance the numbers of O atoms on both sides by adding water:

• Finally we balance the numbers of hydrogen:

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• Example: Balance the following reaction in basic solution.

• Solution: This is a disproportionation reaction; therefore we first add a P4 molecule on the left hand side and balance the stoichiometry.

• Now we determine the change in ON:

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• Therefore,

• Then we balance the number of oxygen atoms:

• And we balance the number of hydrogen atoms:

• Finally we add hydroxide ions on both sides:

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• To obtain the final form of the equation, we combine water and reduce the coefficients into smallest whole numbers:

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• Practice: Balance the following two reactions using the oxidation number method.

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