4.9 - Oxidation-Reduction (Redox) Reactions

Overview

Topic 4.7 showed that a redox reaction moves electrons from the species that is oxidized to the species that is reduced. This topic is about writing balanced equations for those reactions. Redox equations have to balance charge as well as atoms, which makes them hard to balance by guessing. Instead, you split the reaction into two half-reactions, balance each one, and add them back together. You'll use the same half-reactions again in Unit 9 for batteries and electrolysis.

Half-reactions

A half-reaction shows just one side of the electron transfer, with the electrons written in.

In an oxidation half-reaction, electrons appear as a product (they're lost):

\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}}

In a reduction half-reaction, electrons appear as a reactant (they're gained):

\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}

Each half-reaction has to balance on its own, for atoms and for charge. In the zinc half-reaction, the left side has a total charge of 0, and the right side has

(+2) + 2(-1) = 0

The rule that ties the two halves together: electrons lost must equal electrons gained. Electrons don't pile up in a beaker, so a finished redox equation never shows free electrons.

Example 1: Aluminum in a copper(II) solution

When a strip of aluminum is placed in a solution of copper(II) ions, reddish-brown copper metal forms on the aluminum and the blue color of

\mathrm{Cu^{2+}}

fades. Write the balanced net ionic equation.

Step 1: Write the two half-reactions.

\mathrm{Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}} \qquad \text{(oxidation)}

\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)} \qquad \text{(reduction)}

Step 2: Make the electrons equal. Each Al loses 3 electrons and each

\mathrm{Cu^{2+}}

gains 2. The lowest common multiple is 6, so multiply the oxidation by 2 and the reduction by 3.

\mathrm{2Al(s) \rightarrow 2Al^{3+}(aq) + 6e^{-}}

\mathrm{3Cu^{2+}(aq) + 6e^{-} \rightarrow 3Cu(s)}

Step 3: Add them, and cancel the 6 electrons that appear on both sides.

\mathrm{2Al(s) + 3Cu^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s)}

Step 4: Check. There are 2 Al and 3 Cu on each side. The charge is

3(+2) = +6

on the left and

2(+3) = +6

on the right.

4.9 - Oxidation-Reduction (Redox) Reactions

The particle picture shows why the ratio isn't 1:1. Each Al atom gives up three electrons, but each

\mathrm{Cu^{2+}}

ion takes only two, so it takes three copper ions to use up the electrons from two aluminum atoms. A common wrong answer is

\mathrm{Al(s) + Cu^{2+}(aq) \rightarrow Al^{3+}(aq) + Cu(s)}

. Its atoms balance, but its charge doesn't (

+2

on the left,

+3

on the right).

Balancing in acidic solution

Many redox reactions involve ions that contain oxygen, like permanganate (

\mathrm{MnO_4^{-}}

) or dichromate (

\mathrm{Cr_2O_7^{2-}}

), in acid. Their half-reactions need water and

\mathrm{H^{+}}

to balance. Work through each half-reaction in this order:

Balance all atoms except O and H.

Balance O by adding $\mathrm{H_2O}$ .

Balance H by adding $\mathrm{H^{+}}$ .

Balance charge by adding electrons to the more positive side.

Then combine them:

Multiply each half-reaction so the electrons lost equal the electrons gained. Multiply every species in it.

Add the half-reactions and cancel anything that shows up on both sides (the electrons, plus any extra $\mathrm{H_2O}$ or $\mathrm{H^{+}}$ ).

Check that both the atoms and the total charge balance.

Example 2: Permanganate and iron(II)

In acidic solution,

\mathrm{MnO_4^{-}}

oxidizes

\mathrm{Fe^{2+}}

\mathrm{Fe^{3+}}

and is reduced to

\mathrm{Mn^{2+}}

. Balance the equation.

Reduction half-reaction. Mn is already balanced. Add 4

\mathrm{H_2O}

to the right for the 4 O atoms, then 8

\mathrm{H^{+}}

to the left for the 8 H atoms in that water:

\mathrm{MnO_4^{-}(aq) + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)}

The charge is

(-1) + 8(+1) = +7

on the left and

+2

on the right. Add 5 electrons to the left to bring it down to

+2

\mathrm{MnO_4^{-}(aq) + 8H^{+}(aq) + 5e^{-} \rightarrow Mn^{2+}(aq) + 4H_2O(l)}

The 5 electrons match the change in Mn's oxidation number, from

+7

+2

Oxidation half-reaction. Each iron loses one electron, so multiply it by 5 to match the 5 electrons gained:

\mathrm{5Fe^{2+}(aq) \rightarrow 5Fe^{3+}(aq) + 5e^{-}}

Add and cancel the 5 electrons:

\mathrm{MnO_4^{-}(aq) + 5Fe^{2+}(aq) + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)}

Check: 1 Mn, 5 Fe, 4 O, and 8 H on each side. Charge:

(-1) + 5(+2) + 8(+1) = +17

on the left and

(+2) + 5(+3) = +17

on the right.

Example 3: Dichromate and iodide

In acidic solution,

\mathrm{Cr_2O_7^{2-}}

reacts with

\mathrm{I^{-}}

to form

\mathrm{Cr^{3+}}

and

\mathrm{I_2}

. Balance the equation.

Reduction half-reaction. Balance Cr (2

\mathrm{Cr^{3+}}

), then O (7

\mathrm{H_2O}

), then H (14

\mathrm{H^{+}}

). The charge is

(-2) + 14 = +12

on the left and

2(+3) = +6

on the right, so add 6 electrons to the left:

\mathrm{Cr_2O_7^{2-}(aq) + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)}

Each Cr goes from

+6

+3

, and there are two of them:

2 \times 3 = 6

electrons. That matches.

Oxidation half-reaction. Balance I first (2

\mathrm{I^{-}}

make one

\mathrm{I_2}

), then charge:

\mathrm{2I^{-}(aq) \rightarrow I_2(s) + 2e^{-}}

Multiply it by 3 so 6 electrons are lost, then add and cancel:

\mathrm{6I^{-}(aq) \rightarrow 3I_2(s) + 6e^{-}}

\mathrm{Cr_2O_7^{2-}(aq) + 6I^{-}(aq) + 14H^{+}(aq) \rightarrow 2Cr^{3+}(aq) + 3I_2(s) + 7H_2O(l)}

Check: 2 Cr, 7 O, 14 H, and 6 I on each side. Charge:

(-2) + 6(-1) + 14(+1) = +6

on the left and

2(+3) = +6

on the right.

Extra: balancing in basic solution

The CED doesn't ask for this method, but textbooks and labs use it. In basic solution there's essentially no free

\mathrm{H^{+}}

. Balance the equation as if it were acidic, then add one

\mathrm{OH^{-}}

to both sides for each

\mathrm{H^{+}}

. On the side with

\mathrm{H^{+}}

, each

\mathrm{H^{+}}

and

\mathrm{OH^{-}}

combine into

\mathrm{H_2O}

. Then cancel any water that appears on both sides.

For example, permanganate oxidizes sulfite to sulfate in basic solution and is reduced to

\mathrm{MnO_2}

. Balanced as if acidic (6 electrons transferred), the equation is

\mathrm{2MnO_4^{-}(aq) + 3SO_3^{2-}(aq) + 2H^{+}(aq) \rightarrow 2MnO_2(s) + 3SO_4^{2-}(aq) + H_2O(l)}

Add 2

\mathrm{OH^{-}}

to both sides. The 2

\mathrm{H^{+}}

and 2

\mathrm{OH^{-}}

on the left become 2

\mathrm{H_2O}

, and one of them cancels with the

\mathrm{H_2O}

on the right:

\mathrm{2MnO_4^{-}(aq) + 3SO_3^{2-}(aq) + H_2O(l) \rightarrow 2MnO_2(s) + 3SO_4^{2-}(aq) + 2OH^{-}(aq)}

Check: 2 Mn, 3 S, 18 O, and 2 H on each side. Charge:

2(-1) + 3(-2) = -8

on the left and

3(-2) + 2(-1) = -8

on the right.

Counting electrons transferred

The total number of electrons transferred in a balanced redox equation is the number that canceled when you added the half-reactions. It's 6 for aluminum and copper(II), 5 for permanganate and iron(II), and 6 for dichromate and iodide. You'll call this number

n

in Unit 9, where it links a reaction to its cell potential and free energy (

\Delta G^\circ = -nFE^\circ

on the reference sheet). You can always double-check

n

with oxidation numbers: in the dichromate reaction, 2 Cr atoms each drop by 3 (6 electrons gained), and 6 I atoms each rise by 1 (6 electrons lost).

Example 4: Iron in a supplement tablet (AP-style)

An iron supplement tablet is dissolved in dilute sulfuric acid, so all of its iron is present as

\mathrm{Fe^{2+}}

. The solution is titrated with 0.01000 M

\mathrm{KMnO_4}

, and 23.28 mL is needed to reach the endpoint. The reaction is the one balanced in Example 2:

\mathrm{MnO_4^{-}(aq) + 5Fe^{2+}(aq) + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)}

(a) Calculate the mass of iron in the tablet, in milligrams. (b) How many moles of electrons were transferred during the titration?

(a) Moles of permanganate, then the 1:5 mole ratio, then mass:

n_{\mathrm{MnO_4^-}} = (0.02328\ \mathrm{L})(0.01000\ \mathrm{mol/L}) = 2.328 \times 10^{-4}\ \mathrm{mol}

n_{\mathrm{Fe^{2+}}} = 2.328 \times 10^{-4}\ \mathrm{mol\ MnO_4^-} \times \frac{5\ \mathrm{mol\ Fe^{2+}}}{1\ \mathrm{mol\ MnO_4^-}} = 1.164 \times 10^{-3}\ \mathrm{mol}

m = (1.164 \times 10^{-3}\ \mathrm{mol})(55.85\ \mathrm{g/mol}) = 0.06501\ \mathrm{g} = 65.01\ \mathrm{mg}

The tablet contains 65.01 mg of iron, which matches the 65 mg printed on a typical label.

(b) Each

\mathrm{Fe^{2+}}

loses one electron, so the electrons transferred equal the moles of iron:

1.164 \times 10^{-3}

mol of electrons. The other half agrees: each

\mathrm{MnO_4^{-}}

gains 5 electrons, and

5(2.328 \times 10^{-4}) = 1.164 \times 10^{-3}

mol.

Common mistakes

Balancing atoms but not charge. Add up the charges on each side every time.

Putting electrons on the wrong side. Oxidation half-reactions have electrons as products, and reduction half-reactions have them as reactants.

Leaving electrons in the final equation. If they don't cancel, the multipliers are wrong.

Balancing O with O atoms or $\mathrm{O^{2-}}$ . In acid, balance O with $\mathrm{H_2O}$ and H with $\mathrm{H^{+}}$ .

Scaling only part of a half-reaction. When you multiply by 3, multiply every species, including $\mathrm{H^{+}}$ and $\mathrm{H_2O}$ .

Forgetting to cancel. After adding, remove anything that appears on both sides, like extra $\mathrm{H_2O}$ or $\mathrm{H^{+}}$ .

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