Balancing Redox Equations and Titrations: Quick Revision

A brief note on Balancing Redox Equations and Titrations. Read it before a test. Try each question at the end before you open its answer.

The oxidation number method

• Electrons lost equal electrons gained: the total rise in oxidation number equals the total fall.
• Find the change for one atom, then match the changes with the smallest numbers that both divide into.
• Then balance the atoms that did not change, charge, hydrogen and oxygen.
• Only coefficients change. Never change a subscript.

The half-reaction method

• In acid: balance atoms, then oxygen with $\text{H}_2\text{O}$, hydrogen with $\text{H}^+$, charge with electrons.
• In base: balance as in acid, add $\text{OH}^-$ to both sides, combine with $\text{H}^+$ into water, and cancel.
• Make the electrons equal, add the halves, cancel, and check atoms and charge.
• A disproportionation has one element both oxidised and reduced.

Redox titrations

• Permanganate is its own indicator. Use dilute sulfuric acid, never hydrochloric or nitric acid.
• Dichromate needs an indicator. Iodine is shown by starch, and thiosulfate turns it colourless.
• Moles are concentration in mol/L times volume in litres. Take the ratio from the balanced equation.
• n-factors: $\text{KMnO}_4$ in acid 5, $\text{K}_2\text{Cr}_2\text{O}_7$ 6, $\text{Fe}^{2+}$ 1, oxalic acid 2.
• At the end point, $n_1 M_1 V_1 = n_2 M_2 V_2$.

A worked example

A $20.0\ \text{mL}$ portion of oxalic acid needs $17.2\ \text{mL}$ of $0.0150\ \text{mol/L}$ permanganate in acid.

1. Equivalents of permanganate. $5 \times 0.0150 \times 0.0172 = 0.00129$.
2. Equivalents of oxalic acid. The same, $0.00129$. Its n-factor is 2.
3. Concentration. $\dfrac{0.00129}{2 \times 0.0200\ \text{L}} = 0.0322\ \text{mol/L}$.

Practice with new numbers

A program to try

The program does the worked example with equivalents.

Mistakes to avoid

• The mole ratio. One permanganate ion reacts with five iron(II) ions, not one.
• Volumes in mL. Divide by 1000 to get litres.
• Acid choice. Sulfuric acid only for permanganate.
• Charge and water. Water has no charge. Use $\text{H}^+$ or $\text{OH}^-$.
• Electrons left over. They must cancel in the final equation.

Check yourself

1. In acid, the sulfate ion becomes $\text{SO}_2$. How many electrons are gained?

2. What is the n-factor of $\text{K}_2\text{Cr}_2\text{O}_7$ in acid?

3. In base, you have $4\ \text{H}^+$ in a half-reaction balanced as in acid. What do you add?

4. Why is hydrochloric acid not used in a permanganate titration?

5. A titration uses $0.000300\ \text{mol}$ of permanganate for iron(II). How many moles of iron(II) were there?