Balancing redox equations by ion-electron method

1 Balancing redox equations by ion-electron method involves two steps:

(a) Balancing half-equations for oxidation and reduction reactions (b) Combining two half-equations to form a redox equation 2 Writing redox equations for reactions in acidic solution

As an example, consider the reaction involving ClO₃–

(aq) and I–

(aq) ions in acidic conditions.

(a) Write half-equations for the oxidation and reduction processes. Step 1 Determine the reaction products and write two incomplete half-equations. One of them involves an oxidising agent and the other a reducing agent.

I–

(aq) → I2(aq) ...(5)

ClO₃–

(aq) → Cl–

(aq) ...(6)

Step 2 Balance the half-equations in terms of the number of atoms.

(i) Multiply the molecule or ion with a suitable number so that the number of atoms on both sides of the equation are the same.

2I–

(aq) → I2(aq) ...(7)

(ii) Calculate the total number of oxygen atoms on both sides of the equation and add water molecules to the side that has fewer oxygen atoms. Thus, equation (6) becomes

ClO3

–

(aq) → Cl–

(aq) + 3H2O(l) ...(8)

(iii) Add H+

(aq) ions to the side that has fewer hydrogen atoms. Thus, equation (8) becomes

ClO₃–

(aq) + 6H+

(aq) → Cl–

(aq) + 3H₂O(l) ...(9) Step 3 Balance the half-equations in terms of number of charges

by adding electrons. In equation (9), the oxidation number of Cl changes by –6 units. So add 6 electrons for the reduction process. ClO₃–

(aq) + 6H+

(aq) + 6e–→

Cl–

(aq) + 3H₂O(l) ...(10) In equation (7), the oxidation number of iodine increases by +1 unit, so there is a loss of one electron per atom of iodine in the oxidation process. Thus,

2I–

(aq) → I₂(aq) + 2e–

...(11) (2e–

because 2I–

(aq) ions are involved)

Chapter-01.indd 30

Equations (10) and (11) are now balanced in terms of the number of atoms and the number of charges.

(b) Combine the two half-equations to form a redox equation. Step 1 Multiply each half-equation with a suitable factor so that

the number of electrons accepted in one half-equation is the same as the number of electrons donated in the other half-equation. For example, to balance the number of electrons in equation (10) and (11), multiply equation(11) by 3 to obtain:

6I–

(aq) → 3I2(aq) + 6e –

...(12)

Step 2 Combine equations (10) and (12) to give a balanced overall equation that does not contain electrons.

ClO₃– (aq) + 6H+ (aq) + 6e– → Cl– (aq) + 3H₂O(l) 6I– (aq) → 3I2(aq) + 6e – ClO3 – (aq) + 6H+ (aq) + 6I– (aq) → Cl–

(aq) + 3H2O(l) + 3I2(aq)

EXAMPLE 1.26

Balance the following ionic equation. Cr₂O₇2–_{(aq) + H}+_{(aq) + NO} 2 –_{(aq) → Cr}3+_{(aq) + NO} 3 –_(aq) SOLUTION

Step 1 Write half-equations for the oxidation and reduction processes. Cr₂O₇2–_{(aq) + 14H}+_{(aq) + 6e}–→ _2Cr3+_{(aq) + 7H}

2O(l) ...(1) NO₂–_{(aq) + H}

2O(l) → NO3

–_{(aq) + 2H}+_{(aq) + 2e}– _...(2) Step 2 Multiply equation (2) by 3, so that the number of electrons liberated is

the same as the number of electrons received in equation (1). 3NO₂–_{(aq) + 3H}

2O(l) → 3NO3

–_{(aq) + 6H}+_{(aq) + 6e}– _...(3) Step 3 Combine the two half-equations (1) and (3) to form an ionic equation.

Cancel the electrons, hydrogen ions and water molecules that appear on both sides of the equation.

Cr₂O₇2–_{(aq) + 14H}+_{(aq) + 6e}– → _2Cr3+_{(aq) + 7H} 2O(l) 3NO₂–_{(aq) + 3H}

2O(l) → 3NO3

–_{(aq) + 6H}+_{(aq) + 6e}– Cr₂O₇2–_{(aq) + 8H}+_{(aq) + 3NO}

2

–_(aq) → _2Cr3+_{(aq) + 4H}

2O(l) + 3NO3 –_(aq)

3 Writing redox equations for reactions in alkaline solution

The method is the same as for reactions in acidic solution except for one additional step at the end. In the last step, the equation can be balanced by adding H2O(l) or OH

–

(aq). Consider the reaction between Cl₂ and Cr3+

ions in alkaline solution to form CrO4

2–

and Cl–

ions.

(a) Write half-equations for the oxidation and reduction processes using the method described above for reactions in acidic media.

Cr3+ + 4H₂O → CrO₄2– + 8H+ + 3e– ...(1) oxidation Cl2 + 2e – → 2Cl– ...(2) reduction 8 4 Chapter-01.indd 31 Chapter-01.indd 31 4/30/2012 7:11:37 PM4/30/2012 7:11:37 PM

(b) Combine the two half-equations to form a redox equation. Step 1 Multiply equation (1) by 2 and multiply equation (2) by 3 so

that the number of electrons liberated in equation (1) is the same as the number of electrons received in equation (2).

2Cr3+ + 8H₂O → 2CrO₄2– + 16H+ + 6e– ...(3) 3Cl2 + 6e –→ 6Cl– ...(4) Step 2 Combine the two half-equations to form an ionic equation.

2Cr3+

+ 8H2O + 3Cl2→ 2CrO4 2–

+ 16H+

+ 6Cl–

Step 3 Add 16OH–

to both sides of the equation to eliminate H+

ions. Then cancel out the water molecules on both sides of the equation. 2Cr3+ + 8H2O + 3Cl2 + 16OH –→ 2CrO4 2– + 16H+ + 16OH– + 6Cl– 16H₂O That is, 2Cr3+ + 3Cl2 + 16OH –→ 2CrO4 2– + 8H2O + 6Cl –

1. State the oxidation numbers of the highlighted elements in each of the following compounds.

(a) H₂ O₂ (d) NO (g) KIO₃ (b) H₂ O (e) S₂O₃2– _{(h) ICl} (c) NO₃– _{(f) S} 4O6 2– _{(i) F} 2O

2. Identify the balanced equations in the list below. If an equation is not balanced, change the coefficients to balance the equation.

(a) N₂ + 3H₂→ 2NH₃ (d) Pb(NO₃)₂ + K₂CrO₄→ PbCrO₄ + 2KNO₃ (b) 2Al + Fe₂O₃→ Al₂O₃ + 2Fe (e) CH₄ + O₂→ CO₂ + H₂O

(c) 4NO₂ + H₂O + 2O₂→ 4HNO₃ 3. Balance the following equations.

(a) Fe + Cl₂→ FeCl₃ (d) H+ _{+ CO} 3 2–→ _CO 2 + H2O (b) C₃H₈ + O₂→ CO₂ + H₂O (e) I₂ + Na₂S₂O₃→ NaI + Na₂S₄O₆ (c) NH₄NO₃→ N₂ + H₂O + O₂

4. Write (i) the half-equations and (ii) the ionic equations for the following reactions. (a) PbO₂(s) + H+_{(aq) + Cl}–_(aq)

(b) H₂O₂(aq) + MnO₄–_{(aq) + H}+_(aq) (c) Cr₂O₇2–_{(aq) + H}+_{(aq) + Sn}2+_(aq) (d) ClO–_{(aq) + OH}–_{(aq) + CrO}

2

–_{(aq) [Hint: CrO} 2 – _{is oxidised to CrO} 4 2–_.]

QUICK CHECK 1.5

Limiting Reactant

1 It is possible to calculate from a balanced chemical equation, the exact quantities of reactants which are consumed and the products which are formed. However, many reactions are carried out using an excess amount of one reactant.

8

Chapter-01.indd 32

Table 1.8 Results of experiments using different molar ratios of hydrogen and chlorine

Experiment

Amount of reactants (mol) Amount of products (mol) Hydrogen Chlorine Hydrogen

chloride Unreacted hydrogen Unreacted chlorine 1 1.0 1.0 2.0 0 0 2 2.0 1.0 2.0 1.0 0 3 3.0 1.0 2.0 2.0 0

2 Consider the reaction between hydrogen and chlorine to form hydrogen chloride. The chemical equation for the reaction is

H2(g) + Cl2(g) → 2HCl(g)

Table 1.8 shows the results of three experiments obtained by using different molar ratios of the reactants.

(a) In Experiment 1, the ratio of the number of moles of reactants used is the same as that given in the balanced chemical equation. (b) In Experiments 2 and 3, the ratio of the number of moles of

reactants used is different from that given in the balanced chemical equation.

3 From the results shown in Table 1.8, you will notice that:

(a) In Experiment 2, one mole of hydrogen remained unreacted when all the chlorine has been consumed. The reactant which reacts completely in a reaction is known as the limiting reactant or limiting reagent. This is because it determines or limits the amount of product formed. The other reactants are called excess reactants or excess reagents.

(b) Thus, in Experiments 2 and 3, chlorine acts as the limiting reagent because it limits the amount of hydrogen chloride produced to two moles, regardless of the amount of hydrogen used. Once all the chlorine has been consumed, the reaction stops. (c) Hydrogen is the excess reagent and some is left over at the end of

the reaction. Thus, the quantity of products formed in a reaction is always determined by the quantity of the limiting reagent. EXAMPLE 1.27

Zinc reacts with hydrochloric acid according to the equation Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

If 0.05 mole of zinc was added to 0.075 mole of hydrochloric acid, (a) identify the limiting reagent,

(b) calculate the amount (in moles) of zinc chloride formed.

SOLUTION

(a) According to the equation,

0.05 mole of zinc will react with 0.10 mole of HCl (but only 0.075 mole of hydrochloric acid is present).

A limiting reactant • determines the amount

of product formed • yields the lower

amount of product

Chapter-01.indd 33

0.075 mole of HCl will react with 0.0375 mole of zinc (0.05 mole of zinc is present).

Hence, zinc is the excess reagent and HCl is the limiting reagent.

(b) The quantity of products formed in a reaction is determined by the quantity of the limiting reactant.

According to the equation, 2.0 moles HCl will produce 1.0 mole ZnCl₂. That is, 1.0 mole HCl will produce 1

2 mole ZnCl2. 0.075 mole HCl will produce 1

2 × 0.075 = 0.0375 mole ZnCl2. Alternative method

If zinc (0.05 mol) is the limiting reagent, number of moles of ZnCl₂ produced = 0.05 mol If HCl (0.075 mol) is the limiting reagent, number of moles of ZnCl₂ produced = 1

2 × 0.075 mol = 0.0375 mol

Hence, HCl is the limiting reagent because it yields the lower amount of product (0.0375 mole of ZnCl₂).