Redox equations

We calculate the oxidation numbers of all atoms. We find that the oxidation numbers of manganese and bromine change:

Br⁰₂ + Mn^IIBr^–I ₂ + K^IBr^–I + H₂O^–II

We write the partial (half) reactions for oxidation and reduction:

Br⁰ (oxidation)

Mn^II (reduction)

Since the number of electrons gained and lost must be equal, we multiply the first half-reaction by five and add the partial reactions:

5Br⁰

Mn^II                                                       

5Br^–I – 5e^– + Mn^VII + 5e^–= 5Br⁰ + Mn^II

and after simplification we obtain SRR:

5Br⁰ + Mn^II

On the right-hand side of the equation we must have, according to SRR, five bromine atoms with oxidation number 0. However, bromine exists as diatomic molecules, so the stoichiometric coefficient would be fractional. Therefore we multiply SRR by 2:

10Br^–I + 2Mn^VII → 10Br⁰ + 2Mn^II

The numerical values we obtained are not the stoichiometric coefficients but express the numbers of atoms that must appear on the left and right sides of the chemical equation. On both sides we must have two manganese atoms, so the coefficient before KMnO₄ and MnBr₂ will be 2:

2KMnO₄ + HBr → Br₂ + 2MnBr₂ + KBr + H₂O

At the same time, on the left-hand side we must have ten bromine atoms with oxidation number –I, but we cannot put coefficient 10 before HBr because on the right-hand side bromine also appears with unchanged oxidation number –I. Thus the coefficient before HBr must be larger than 10. On the right-hand side we must have 10 bromine atoms with oxidation number 0, therefore we put coefficient 5 before Br₂:

2KMnO₄ + HBr → 5Br₂ + 2MnBr₂ + KBr + H₂O

On the left-hand side we have two potassium atoms, therefore we place coefficient 2 before KBr:

2KMnO₄ + HBr → 5Br₂ + 2MnBr₂ + 2KBr + H₂O

Since we now have coefficients before all bromine-containing substances on the right-hand side, we find that there are 16 bromine atoms on the right, therefore we put coefficient 16 before HBr:

2KMnO₄ + 16HBr → 5Br₂ + 2MnBr₂ + 2KBr + H₂O

The coefficient 16 before HBr also means that 10 bromine atoms participated in the redox reaction and the remaining six remained unchanged. On the left-hand side there are also 16 hydrogen atoms, therefore we put coefficient 8 before H₂O on the right:

→ 5Br₂ + 2MnBr₂ + 2KBr + 8H₂O

In this way we obtained the stoichiometric coefficients for all substances in the chemical equation. We verify the solution by counting oxygen atoms on both sides of the equation (8 = 8). Since the obtained stoichiometric coefficients (2, 16 = 5, 2, 2, 8) have no common divisor other than 1, the balancing of the chemical equation is complete.

We calculate the oxidation numbers of all atoms. We find that only the oxidation number of the rhenium atom changes:

Re^VIF^–I₆ + H^I₂O^–II → H^IRe^VIIO^–II₄ + Re^IVO^–II₂ + H^IF^–I

This is a disproportionation equation. We write the partial (half) reactions for oxidation and reduction:

Re^VII (oxidation)

→ Re^IV (reduction)

Since the number of electrons gained and lost must be equal, we multiply the first half-reaction by two and add the partial reactions:

2Re^VI – 2e^– → 2Re^VII

Re^VI + 2e^– → Re^IV                                                          

2Re^VI – 2e^– + Re^VI + 2e^– = 2Re^VII + Re^IV

After simplification we obtain SRR:

2Re^VII + Re^IV

From SRR it follows that on the left-hand side we must have three rhenium atoms with oxidation number VI and on the right-hand side two rhenium atoms with oxidation number VII and one atom with oxidation number IV. We add stoichiometric coefficients to the redox equation:

3ReF₆ + H₂O → 2HReO₄ + ReO₂ + HF

On the left-hand side we have 18 fluorine atoms, therefore we put coefficient 18 before HF:

3ReF₆ + H₂O → 2HReO₄ + ReO₂ + 18HF

The only substance without a stoichiometric coefficient is water. On the right-hand side we have 2 + 18 = 20 hydrogen atoms, therefore we put coefficient 10 before H₂O:

2HReO₄ + ReO₂ + 18HF

Calculate the oxidation numbers of all atoms. We find that only the oxidation number of the manganese atom changes:

(Mn^VIO^–II₄)^2– + (H^I₃O^–II)⁺ → (Mn^VIIO^–II₄)^– + Mn^IVO^–II₂ + H^I₂O^–II

This is a disproportionation equation. We write the partial oxidation and reduction equations:

Mn^VI – 1e^– → Mn^VII (oxidation)

Mn^VI + 2e^– → Mn^IV (reduction)

Since the number of electrons accepted and released must be the same, we multiply the first partial equation by two and add the partial equations:

2Mn^VI – 2e^– → 2Mn^VII

2Mn^VI – 2e^– + Mn^VI + 2e^– → 2Mn^VII + Mn^IV

and after adjustment we obtain the net ionic result:

→ 2Mn^VII + Mn^IV

From this it follows that on the left side of the equation we must have three manganese atoms with oxidation number VI, and on the right side two manganese atoms with oxidation number VII and one atom with oxidation number IV. We add stoichiometric coefficients to the redox equation:

3MnO₄^2– + H₃O⁺ → 2MnO₄^– + MnO₂ + H₂O

Two species in the equation remain without coefficients: the reactant—oxonium cation—and the product—water. Since this is an ionic equation, we use charge balance. On the right side only one ion has a coefficient, so we can compute the sum of charges: 2 · (–1) + 1 · 0 + 1 · 0 = –2. On the left side we must also have a total charge of –2, but we currently have (besides H₃O⁺): 3 · (–2) = –6. We need to increase this sum by 4, therefore the coefficient in front of H₃O⁺ will be 4:

3MnO₄^2– + 4H₃O⁺ → 2MnO₄^– + MnO₂ + H₂O

On the left we have 12 hydrogen atoms, so we put the coefficient 6 in front of H₂O:

3MnO₄^2– + 4H₃O⁺ → 2MnO₄^– + MnO₂ + 6H₂O

Calculate the oxidation numbers of all atoms. We find that only the oxidation number of the nitrogen atom changes; we underline its symbols:

N^IVO^–II₂ + N^IIO^–II + Na^IO^–IIH^I → Na^IN^IIIO^–II₂ + H^I₂O^–II

This is a comproportionation equation. We write the partial oxidation and reduction equations:

N^II – 1e^– → N^III (oxidation)

N^IV + 1e^– → N^III (reduction)

Since the number of electrons accepted and released is already the same, we add the partial equations:

N^III

→ N^III                                               

N^II – 1e^– + N^IV + 1e^– →N^III + N^III

and after adjustment we obtain the net ionic result:

N^II + N^IV → 2N^III

From this it follows that on the right side of the equation we must have two nitrogen atoms with oxidation number III, and on the left one nitrogen atom with oxidation number II and one with IV. We add stoichiometric coefficients to the redox equation:

NO₂ + NO + NaOH → 2NaNO₂ + H₂O

On the right side there are two sodium atoms, so we place the coefficient 2 in front of NaOH:

NO₂ + NO + 2NaOH → 2NaNO₂ + H₂O

The only substance without a stoichiometric coefficient is water. On the left side we have 2 hydrogen atoms, so the coefficient in front of H₂O must be 1:

NO₂ + NO + 2NaOH → 2NaNO₂ + H₂O

We calculate the oxidation numbers of all atoms. We find that only the oxidation numbers of the iodine atoms change; we underline their symbols:

→ I⁰₂ + H^I₂O^–II

This is a symproportionation (comproportionation) equation. We write the half-reactions for oxidation and reduction:

→ I⁰ (oxidation)

I^V + 5e^– → I⁰ (reduction)

Because the number of electrons accepted and donated must be equal, we multiply the first half-reaction by five and add the half-reactions:

5I^–I – 5e^– → 5I⁰

I^V + 5e^– → I⁰

5I^–I – 5e^– + I^V + 5e^– → 5I⁰ + I⁰

After simplification we obtain the net atom-reconciliation relation (ARR):

5I^–I + I^V → 6I⁰

The numerical values we obtained are not the stoichiometric coefficients but express the numbers of atoms that must appear on the left and right sides of the chemical equation. From the ARR it follows that the left side must contain five iodine atoms with oxidation number –I, so we place coefficient 5 before I^–. Also, there must be one iodine atom with oxidation number V on the left, so the coefficient before IO₃^– is 1. On the right side there must be six iodine atoms with oxidation number 0, therefore we put coefficient 3 before I₂:

5I^– + IO₃^– + H₃O⁺ → 3I₂ + H₂O

Two species in the equation still lack coefficients: the reactant oxonium cation and the product water. Since this is an ionic equation, we use the rule of charge balance. On the right side there are no ions, so the sum of charges on the right is zero. The same total charge must be on the left. So far (excluding H₃O⁺) we have 5·(–1) + 1·(–1) = –6. We need to increase this sum by +6, therefore the coefficient before H₃O⁺ will be 6:

5I^– + IO₃^– + 6H₃O⁺ → 3I₂ + H₂O

There are 18 hydrogen atoms on the left, so we put coefficient 9 before H₂O:

5I^– + IO₃^– + 6H₃O⁺ → 3I₂ + 9H₂O

With this procedure we obtained the stoichiometric coefficients for all species in the chemical equation. We verify the solution by counting oxygen atoms on both sides (9 = 9). Since the obtained coefficients (5, 1, 6 = 3, 9) have no common divisor other than 1, balancing the chemical equation is complete.

We calculate the oxidation numbers of all atoms. We find that the oxidation numbers of all atoms change: iron, sulfur, and oxygen:

Fe^IIS^–I₂ + O⁰₂ → Fe^III₂O^–II₃ + S^IVO^–II₂

We write the half-reactions for oxidation and reduction:

Fe^II – 1e^– → Fe^III (oxidation)

S^–I – 5e^– → S^IV (oxidation)

O⁰ + 2e^– → O^–II (reduction)

Since iron and sulfur occur together in the same compound on the left, their mutual ratio indicated by stoichiometric indices must be preserved. Therefore we multiply the first half-reaction by one, the second half-reaction by two and add them:

Fe^II – 1e^– → Fe^III

2S^–I – 10e^– → 2S^IV

Fe^II – 1e^– + 2S^–I – 10e^– → Fe^III + 2S^IV

After adjustment we obtain the combined half-reaction:

Fe^II + 2S^–I – 11e^– → Fe^III + 2S^IV

We will now adjust this combined half-reaction together with the half-reaction for oxygen:

Fe^II + 2S^–I – 11e^– → Fe^III + 2S^IV

O⁰ + 2e^– →O^–II

Because the number of electrons accepted and donated must be equal, we multiply the first combined half-reaction by 2 and the oxygen half-reaction by 11 and add them:

2Fe^II + 4S^–I – 22e^– → 2Fe^III + 4S^IV

11O⁰ + 22e^– → 11O^–II

2Fe^II + 4S^–I – 22e^– + 11O⁰ + 22e^– → 2Fe^III + 4S^IV + 11O^–II

After simplification we obtain the ARR:

2Fe^II + 4S^–I + 11O⁰ → 2Fe^III + 4S^IV + 11O^–II

The left side must contain eleven oxygen atoms with oxidation number 0. Since oxygen is present as diatomic molecules, the stoichiometric coefficient would be fractional. Therefore we multiply the ARR by 2:

4Fe^II + 8S^–I + 22O⁰ → 4Fe^III + 8S^IV + 22O^–II

The left side should contain 4 iron atoms and 8 sulfur atoms, so we put coefficient 4 before FeS₂. We also need 22 oxygen atoms on the left, so we put coefficient 11 before O₂:

4FeS₂ + 11O₂ → Fe₂O₃ + SO₂

On the right side there must also be four iron atoms and eight sulfur atoms, therefore we put coefficient 2 before Fe₂O₃ and coefficient 8 before SO₂:

4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

With this procedure we obtained the stoichiometric coefficients for all species in the chemical equation. We verify the solution by counting oxygen atoms on both sides (22 = 22). Since the coefficients (4, 11 = 2, 8) have no common divisor other than 1, balancing is complete.

We calculate the oxidation numbers of all atoms. Since the cyanide anion essentially remains unchanged during the reaction (only coordinated), we do not need to determine oxidation numbers of carbon and nitrogen separately; it is sufficient to consider the charge of the whole CN^– anion. We find that the oxidation numbers of gold and oxygen change.

Au⁰ + K^ICN^– + O⁰₂ + H^I₂O^–II → K^I[Au^I(CN^–)₂] + K^IO^–IIH^I

We write the half-reactions for oxidation and reduction:

Au⁰ – 1e^– → Au^I (oxidation)

O⁰ + 2e^– → O^–II (reduction)

Because the number of electrons accepted and donated must be equal, we multiply the first half-reaction by two and add the half-reactions:

2Au⁰ – 2e^– → 2Au^I

O⁰ + 2e^– → O^–II

2Au⁰ – 2e^– + O⁰ + 2e^– → 2Au^I + O^–II

and after simplification we obtain the ARR:

2Au⁰ + O⁰ → 2Au^I + O^–II

The left side must contain two oxygen atoms in oxidation state 0, but oxygen is diatomic, so the stoichiometric coefficient would be fractional. Therefore we multiply the ARR by 2:

4Au⁰ + 2O⁰ → 4Au^I + 2O^–II

The numerical values obtained express numbers of atoms that must be on the left and right. Both sides must contain four gold atoms, so the coefficient before Au and before K[Au(CN)₂] will be 4.

4Au + KCN + O₂ + H₂O → 4K[Au(CN)₂] + KOH

Also, the left side must contain two oxygen atoms in oxidation state 0, which is already satisfied by coefficient 1 before O₂. We cannot place coefficient 2 before KOH on the right because it contains not only the oxygen atoms participating in the redox reaction but also those coming from the water reactant. Since there are eight cyanide anions on the right, we place coefficient 8 before KCN:

4Au + 8KCN + O₂ + H₂O → 4K[Au(CN)₂] + KOH

There are eight potassium atoms on the left; on the right four potassium atoms are bound in 4K[Au(CN)₂], therefore we put coefficient 4 before KOH:

4Au + 8KCN + O₂ + H₂O → 4K[Au(CN)₂] + 4KOH

The last species without a stoichiometric coefficient is water. Since there are four hydrogen atoms on the right, we put coefficient 2 before H₂O:

4Au + 8KCN + O₂ + 2H₂O → 4K[Au(CN)₂] + 4KOH

With this procedure we obtained stoichiometric coefficients for all species in the chemical equation. We verify the solution by counting oxygen atoms on both sides (4 = 4). Since the coefficients (4, 8, 1, 2 = 4, 4) have no common divisor other than 1, balancing is complete.