Lesson 2 of 4Electrochemistry

Unit 5 · Lesson 2 of 4

Balancing Redox with Half-Reactions

Splitting a redox reaction into an oxidation half-reaction and a reduction half-reaction, balancing each for atoms and charge, and combining them. Includes the extra steps for reactions in acidic and in basic solution.

3 min read9 practice questionsAsk AI about this →

Key things to remember

1Order in acidic solution: other atoms, then O with $H_{2} O$ , then H with $H^{+}$ , then charge with $e^{-}$ .
2Oxidation half-reaction: electrons are products. Reduction half-reaction: electrons are reactants.
3Multiply the half-reactions so the electrons lost equal the electrons gained, then add and cancel.
4In basic solution, add $OH^{-}$ to both sides to match each $H^{+}$ , form water, and cancel.
5A correctly balanced redox equation balances in both atoms and total charge.

Many redox reactions happen in water, where $H_{2} O$ , $H^{+}$ or $OH^{-}$ take part but are often left out of the skeleton equation you are given. Try balancing $Cr_{2} O_{7}^{2 -} + Fe^{2 +} \to Cr^{3 +} + Fe^{3 +}$ by inspection and you will get stuck, because the oxygen has nowhere to go. The half-reaction method solves this by splitting the reaction into two halves: one showing only the oxidation, one showing only the reduction. You balance each half completely, including its electrons, then add them back together.

Half-reactions are more than a balancing trick. In the next topic you will see that the two halves can take place in two separate beakers joined by a wire. That is a battery. Tables of standard reduction potentials are tables of half-reactions.

The Method in Acidic Solution

Write the two skeleton half-reactions. Use oxidation numbers if you are unsure which is oxidation and which is reduction.
In each half, balance all atoms other than O and H.
Balance O by adding $H_{2} O$ to the side that needs oxygen.
Balance H by adding $H^{+}$ to the side that needs hydrogen.
Balance charge by adding electrons ( $e^{-}$ ) to the more positive side. Electrons appear on the product side for oxidation and on the reactant side for reduction.
Multiply each half-reaction so both involve the same number of electrons.
Add the halves and cancel anything that appears on both sides (electrons always cancel completely; some $H_{2} O$ and $H^{+}$ may too).
Check that atoms and total charge both balance.

Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O

A balanced reduction half-reaction. Charge: left -2 + 14 - 6 = +6; right 2(+3) = +6.

Charge is the final check

A half-reaction can balance in atoms and still be wrong. Add up the charges on each side: they must be equal. The electrons are there precisely to make the charges match.

Extra Steps in Basic Solution

In a basic solution there is almost no $H^{+}$ , so a final equation containing $H^{+}$ would describe a reaction that cannot happen as written. The simplest approach is to balance as if the solution were acidic, then neutralize the $H^{+}$ .

Balance the overall equation using the acidic-solution steps.
Add as many $OH^{-}$ ions as there are $H^{+}$ ions to both sides. Adding the same thing to both sides keeps the equation balanced.
On the side with both, combine them: $H^{+} + OH^{-} \to H_{2} O$ .
Cancel water molecules that appear on both sides, then check atoms and charge again.

Fig 1Common reduction half-reactions ordered by standard reduction potential (approximate typical table values). Every entry is written as a balanced reduction; reverse one to get an oxidation half-reaction.

Never add free oxygen atoms or free electrons to fix a final equation

Oxygen is balanced only with water, hydrogen only with

H^{+}

(or water and

OH^{-}

in base), and electrons must cancel out of the overall equation. If electrons remain after adding the halves, the multipliers in step 6 are wrong.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Balance

Cr_{2} O_{7}^{2 -} + Fe^{2 +} \to Cr^{3 +} + Fe^{3 +}

in acidic solution.

Example 2

Balance

MnO_{4}^{-} + C_{2} O_{4}^{2 -} \to Mn^{2 +} + CO_{2}

in acidic solution.

Example 3

Balance

MnO_{4}^{-} + I^{-} \to MnO_{2} + I_{2}

in basic solution.

Example 4

Balance

MnO_{4}^{-} + CN^{-} \to MnO_{2} + CNO^{-}

in basic solution.

Where marks get lost

Adding electrons to the wrong side. Check by charge: the electrons always go on the more positive side of the half-reaction.
Balancing O with $O_{2}$ or with free O atoms instead of with $H_{2} O$ .
Stopping once the atoms balance and never checking total charge, which is the only way to catch a wrong number of electrons.
In basic solution, adding $OH^{-}$ to only one side, which unbalances the equation.
Leaving electrons in the final overall equation because the half-reactions were not multiplied to the same number of electrons.

Flashcards 7

Half-reaction

Acidic balancing order

Where do electrons go in an oxidation half-reaction?

Converting an acidic answer to basic

+ 3 more

Practice

Which is the correctly balanced half-reaction for

NO_{3}^{-} \to NO_{2}

in acidic solution?

Balance

Cu + NO_{3}^{-} \to Cu^{2 +} + NO_{2}

in acidic solution with the lowest whole-number coefficients. What is the coefficient of

H^{+}

How many electrons must be added to balance

Cr_{2} O_{7}^{2 -} + 14 H^{+} \to 2 Cr^{3 +} + 7 H_{2} O ?

electrons

A student balances a reaction in acidic solution and gets

\dots + 6 H^{+} \to \dots

. To convert it to basic solution, what should they do next?

Balance

H_{2} O_{2} + Fe^{2 +} \to Fe^{3 +} + H_{2} O

in acidic solution with the lowest whole-number coefficients. What is the coefficient of

Fe^{2 +}

Balance

MnO_{4}^{-} + SO_{3}^{2 -} \to MnO_{2} + SO_{4}^{2 -}

in basic solution with the lowest whole-number coefficients. What is the coefficient of

OH^{-}

(it appears as a product)?

Put the steps for balancing a half-reaction in acidic solution in order.

1Balance O by adding H2O
2Balance charge by adding electrons
3Balance H by adding H+
4Balance atoms other than O and H

Match each half-reaction to its description.

Zn \to Zn^{2 +} + 2 e^{-}

Ag^{+} + e^{-} \to Ag

O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O

2 Cl^{-} \to Cl_{2} + 2 e^{-}

When balancing a half-reaction in acidic solution, what species do you add to balance oxygen atoms? Give its formula.

Spot a mistake on this page? Let us know