Reduction and Oxidation Reactions

Lessons

Senior 4 · Chemistry

Reduction and Oxidation Reactions

5 lessons

Lesson 4 of 5

Balancing Redox Equations

30 min

When inspection fails

Try to balance this by trial and error: $\mathrm{MnO_4^- + C_2O_4^{2-} + H^+ \rightarrow Mn^{2+} + CO_2 + H_2O}$ . You can make the atoms balance in several ways that leave the charges unequal, and only one way in which atoms and charge both balance. Redox equations in solution are hard to balance by inspection because water, hydrogen ions or hydroxide ions take part, and because the electrons lost and gained must match. This lesson gives two systematic methods: the half-equation (ion–electron) method and the oxidation-number method.

Definition

Balanced redox equation: an equation in which (1) every element has the same number of atoms on both sides, (2) the total charge is the same on both sides, and (3) the electrons lost in oxidation equal the electrons gained in reduction.

Acidic medium: a solution containing excess $\mathrm{H^+}$ ions; $\mathrm{H^+}$ and $\mathrm{H_2O}$ may be added to balance. Alkaline (basic) medium: a solution containing excess $\mathrm{OH^-}$ ions; $\mathrm{OH^-}$ and $\mathrm{H_2O}$ may be added.

Figure 16.4: The half-equation method. Step 5 is where electrons appear; step 6 makes them cancel.

Method 1: half-equations in acid

Follow Figure 16.4 for manganate(VII) and ethanedioate (oxalate) ions in acid.

Step	Reduction half	Oxidation half
1. Split	$\mathrm{MnO_4^- \rightarrow Mn^{2+}}$	$\mathrm{C_2O_4^{2-} \rightarrow CO_2}$
2. Atoms other than O, H	Mn already balanced	$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$
3. O with $\mathrm{H_2O}$	$\mathrm{MnO_4^- \rightarrow Mn^{2+} + 4H_2O}$	O balanced (4 = 4)
4. H with $\mathrm{H^+}$	$\mathrm{MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O}$	no H
5. Charge with $e^{-}$	$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$ (charge +2 = +2)	$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-}$ (charge −2 = −2)
6. Equal electrons	× 2: 10 $e^{-}$ gained	× 5: 10 $e^{-}$ lost

Step 7 , adding: $\mathrm{2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2}$ Check: Mn 2 = 2; C 10 = 10; O 8 + 20 = 28 and 8 + 20 = 28; H 16 = 16; charge $- 2 + 16 - 10 = + 4$ and $2 \times (+2) = +4$ . Balanced.

Method 1 in alkaline solution

Balance as if in acid, then add as many $\mathrm{OH^-}$ ions to both sides as there are $\mathrm{H^+}$ ions; each $\mathrm{H^+ + OH^-}$ pair becomes $\mathrm{H_2O}$ ; finally cancel water. This is worked in Worked example 1.

Method 2: oxidation numbers

Assign oxidation numbers and find the atoms that change.
Work out the increase per atom (oxidation) and the decrease per atom (reduction).
Choose coefficients so that the total increase equals the total decrease.
Balance the remaining atoms (spectator ions, then O with $\mathrm{H_2O}$ , then H) by inspection.

For copper and concentrated nitric acid, $\mathrm{Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO_2 + H_2O}$ : Cu rises from 0 to +2 (increase 2); N falls from +5 to +4 (decrease 1). So 2 $\mathrm{NO_2}$ are needed for each Cu: $\mathrm{Cu + ?HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + ?H_2O}$ . Nitrogen: 2 in $\mathrm{Cu(NO_3)_2}$ + 2 in $\mathrm{NO_2}$ = 4 $\mathrm{HNO_3}$ . Hydrogen: 4 H → 2 $\mathrm{H_2O}$ . $\mathrm{Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O}$ Check O: 12 = 6 + 4 + 2.

Worked example 1: manganate(VII) and iodide in alkali

In alkaline solution, manganate(VII) oxidises iodide ions to iodate(V) ions and is reduced to manganese(IV) oxide. Balance the equation.

Solution. Reduction (Mn +7 → +4): $\mathrm{MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O}$ . Oxidation (I −1 → +5): $\mathrm{I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-}$ . Multiply the reduction by 2 (6 $e^{-}$ ) and add: $\mathrm{2MnO_4^- + 8H^+ + I^- + 3H_2O \rightarrow 2MnO_2 + 4H_2O + IO_3^- + 6H^+}$ Cancel: $\mathrm{2MnO_4^- + 2H^+ + I^- \rightarrow 2MnO_2 + H_2O + IO_3^-}$ . The solution is alkaline, so add $\mathrm{2OH^-}$ to each side: $\mathrm{2H^+ + 2OH^-}$ become $\mathrm{2H_2O}$ : $\mathrm{2MnO_4^- + I^- + 2H_2O \rightarrow 2MnO_2 + IO_3^- + H_2O + 2OH^-}$ Cancel one water: $\mathrm{2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-}$ Check: O 9 = 9; H 2 = 2; charge −3 = −3.

Worked example 2: dichromate(VI) and iron(II), with a titration

Balance the reaction of acidified dichromate(VI) with iron(II) ions, then calculate the volume of 0.0200 mol dm⁻³ $\mathrm{K_2Cr_2O_7}$ that reacts with 25.0 cm³ of 0.120 mol dm⁻³ $\mathrm{Fe^{2+}}$ .

Solution. Reduction: $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$ . Oxidation: $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$ , × 6. $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$ Charge check: $- 2 + 14 + 12 = + 24$ ; $+ 6 + 18 = + 24$ . $n(\mathrm{Fe^{2+}}) = 0.120 \times 0.0250 = 3.00 \times 10^{-3}\ \mathrm{mol} \qquad n(\mathrm{Cr_2O_7^{2-}}) = \frac{3.00 \times 10^{-3}}{6} = 5.00 \times 10^{-4}\ \mathrm{mol}$ $V = \frac{5.00 \times 10^{-4}}{0.0200} = 0.0250\ \mathrm{dm^3} = 25.0\ \mathrm{cm^3}$

Warning

Common mistakes. (1) Balancing oxygen with $\mathrm{O_2}$ or $\mathrm{O^{2-}}$ : in solution use $\mathrm{H_2O}$ (and $\mathrm{H^+}$ or $\mathrm{OH^-}$ ). (2) Adding $\mathrm{H^+}$ to an equation for an alkaline solution and leaving it there. (3) Adding electrons to the wrong side: count the charge on each side first, then add electrons to the more positive side. (4) Forgetting to check the charge at the end; an equation with balanced atoms but unequal charge is wrong.

Safety. The titration in Worked example 2 uses potassium dichromate(VI), which is toxic and a cause of cancer: use small volumes of dilute solution, a pipette filler (never your mouth), gloves and goggles, and put the waste in the labelled container.

Tip

"O, H, charge": after the main atoms, balance O with water, H with $\mathrm{H^+}$ , then charge with electrons. In alkali, finish by adding $\mathrm{OH^-}$ to both sides to cancel every $\mathrm{H^+}$ .

Professor Dave Explains: Balancing Redox Reactions in Acidic and Basic Conditions (8 min). Watch each step of the half-equation method, and the extra hydroxide step for basic solutions.

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Practice

Balance the half-equation $\mathrm{NO_3^- \rightarrow NO}$ in acid solution.
Balance the half-equation $\mathrm{SO_2 \rightarrow SO_4^{2-}}$ in acid solution.
Use your answers to 1 and 2 to write the overall equation for the reaction of nitrate with sulfur dioxide in acid.
Balance, by the half-equation method, $\mathrm{H_2O_2 + I^- \rightarrow I_2 + H_2O}$ in acid.
Balance $\mathrm{MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}}$ in acid.
Balance, by the oxidation-number method, $\mathrm{Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO + H_2O}$ (dilute acid).
Chlorine reacts with hot sodium hydroxide to give chloride and chlorate(V) ions. Balance the ionic equation $\mathrm{Cl_2 + OH^- \rightarrow Cl^- + ClO_3^- + H_2O}$ .
25.0 cm³ of a solution of ethanedioate ions needs 18.0 cm³ of 0.0200 mol dm⁻³ $\mathrm{KMnO_4}$ in acid. Calculate the concentration of the ethanedioate ions.

Show answersHide answers

$\mathrm{NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O}$ .
$\mathrm{SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-}$ .
× 2 and × 3 (6 electrons): $\mathrm{2NO_3^- + 3SO_2 + 2H_2O \rightarrow 2NO + 3SO_4^{2-} + 4H^+}$ .
$\mathrm{H_2O_2 + 2H^+ + 2I^- \rightarrow I_2 + 2H_2O}$ .
$\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$ .
Cu: +2 per atom; N: −3 per atom (+5 → +2), so 3 Cu for 2 NO: $\mathrm{3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O}$ .
Cl → $\mathrm{Cl^-}$ gains 1 $e^{-}$ ; Cl → $\mathrm{ClO_3^-}$ loses 5 $e^{-}$ : 5 $\mathrm{Cl^-}$ for each $\mathrm{ClO_3^-}$ . $\mathrm{3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O}$ .
$n(\mathrm{MnO_4^-}) = 0.0200 \times 0.0180 = 3.60 \times 10^{-4}\ \mathrm{mol}$ ; ratio 2 : 5, so $n(\mathrm{C_2O_4^{2-}}) = \frac{5}{2} \times 3.60 \times 10^{-4} = 9.00 \times 10^{-4}\ \mathrm{mol}$ ; concentration $= \frac{9.00 \times 10^{-4}}{0.0250} = 0.0360\ \mathrm{mol\,dm^{-3}}$ .

Quick check

Put the steps of the half-equation method (acid solution) in order.

Tap the cards in the right order.

Key points

A balanced redox equation balances atoms, charge, and electrons lost against electrons gained.
Half-equation method: split; balance atoms; O with $\mathrm{H_2O}$ ; H with $\mathrm{H^+}$ ; charge with $e^{-}$ ; equalise electrons; add and cancel. In alkali, convert $\mathrm{H^+}$ to water with $\mathrm{OH^-}$ .
Oxidation-number method: make the total increase in oxidation number equal the total decrease, then balance the rest by inspection.
Key ratios: $\mathrm{MnO_4^- : Fe^{2+} = 1 : 5}$ ; $\mathrm{Cr_2O_7^{2-} : Fe^{2+} = 1 : 6}$ ; $\mathrm{MnO_4^- : C_2O_4^{2-} = 2 : 5}$ .

Sources and credits

Video: Balancing Redox Reactions in Acidic and Basic Conditions, Professor Dave Explains, embedded from YouTube.
Written by Ganzaa: the text, flow chart, tables, worked examples and Practice items.

1When inspection fails

2Method 1: half-equations in acid

3Method 1 in alkaline solution

4Method 2: oxidation numbers

When inspection fails

Method 1: half-equations in acid

Method 1 in alkaline solution

Method 2: oxidation numbers