Redox reactions and oxidation numbers in water: VCE Chemistry Unit 2

What this dot point is asking

VCAA wants you to assign oxidation numbers, identify the species oxidised and reduced (and the corresponding reductant and oxidant), and to balance half-equations and overall ionic equations for redox reactions in acidic solution.

The answer

Oxidation numbers (rules)

The oxidation number (or oxidation state) is the hypothetical charge an atom would have if every bond were ionic. Use the rules in order; later rules yield to earlier ones if there is a conflict.

1. Elements in their standard state: ox. no. = 0 (Na, O2, S8, Hg(l)).
2. Monatomic ions: ox. no. = charge (Na^+ is +1, S^2- is -2).
3. Group 1 metals in compounds: +1. Group 2 metals in compounds: +2.
4. Fluorine in compounds: -1.
5. Hydrogen: usually +1 (in H2O, HCl), but -1 in metal hydrides (NaH, CaH2).
6. Oxygen: usually -2 (in H2O, CO2), but -1 in peroxides (H2O2), -1/2 in superoxides, and +2 in OF2.
7. Sum of oxidation numbers = overall charge of the species (0 for a neutral compound, the ion charge for an ion).

Quick examples:

• MnO4^-: O is -2 (4 x -2 = -8). Total must be -1, so Mn = +7.
• Cr2O7^2-: O is -2 (7 x -2 = -14). 2Cr + (-14) = -2, so Cr = +6 each.
• SO3^2- (sulfite): S = +4.
• CH4: C = -4 (using H = +1, 4 x +1 = +4, so C = -4 for total of 0).
• CH3COOH: average C = 0 (but the C in COOH is +3 and the C in CH3 is -3; the average matters for the overall balance).

Oxidation and reduction

Oxidation is loss of electrons; the oxidation number increases.
Reduction is gain of electrons; the oxidation number decreases.

Memory aid: OIL RIG (Oxidation Is Loss; Reduction Is Gain).

The oxidant (or oxidising agent) accepts electrons; it is itself reduced.
The reductant (or reducing agent) donates electrons; it is itself oxidised.

For Zn + Cu^2+ -> Zn^2+ + Cu: Zn goes 0 to +2 (oxidised, the reductant); Cu^2+ goes +2 to 0 (reduced, the oxidant).

Balancing half-equations in acidic solution

A half-equation is a balanced equation showing one half of the redox reaction with electrons explicitly. Use this procedure for acidic solution:

1. Balance the redox atom (the element whose ox. no. is changing).
2. Balance O by adding H2O to the side that needs O.
3. Balance H by adding H^+ to the side that needs H.
4. Balance charge by adding electrons (e^-) to the side that needs more negative charge.
5. Cancel any species that appear on both sides; simplify.

Example: convert MnO4^- to Mn^2+ in acidic solution.

1. Mn balanced: 1 MnO4^- on left, 1 Mn^2+ on right.
2. O balance: 4 O on left, 0 O on right. Add 4 H2O on the right: MnO4^- -> Mn^2+ + 4H2O.
3. H balance: 8 H on right, 0 on left. Add 8 H^+ on the left: MnO4^- + 8H^+ -> Mn^2+ + 4H2O.
4. Charge balance: left = -1 + 8(+1) = +7. Right = +2. Add 5 e^- on the left: MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O.

The full reduction half is now balanced.

Combining half-equations

To get the overall equation:

1. Multiply each half by the appropriate factor so the electrons cancel (find the LCM of the electrons in each half).
2. Add the two halves.
3. Cancel any species (H^+, H2O) that appear on both sides.
4. Check that atoms and charge balance.

The final equation will not contain any electrons.

Quick checklist before submitting

• Every atom balanced on both sides.
• Every charge balanced on both sides.
• Electrons cancelled in the overall equation (only present in half-equations).
• States included where the question expects them.

Examples in context

Example 1. Dissolved oxygen in Yarra River fish kills. Melbourne Water responds to summer fish kills downstream of Dights Falls when dissolved oxygen drops below $4 \, \text{mg/L}$. Decaying algae consume oxygen via respiration: $\text{C}_6 \text{H}_{12} \text{O}_6 + 6 \text{O}_2 \to 6 \text{CO}_2 + 6 \text{H}_2 \text{O}$, in which carbon goes from oxidation state $0$ in glucose to $+4$ in $\text{CO}_2$ (oxidation) and oxygen from $0$ to $-2$ (reduction). Operators measure DO with a Clark-cell electrode based on the half-equation $\text{O}_2 + 4 \text{H}^+ + 4 e^- \to 2 \text{H}_2 \text{O}$. Treatment includes mechanical aeration that re-dissolves atmospheric oxygen, restoring DO from $2$ to $7 \, \text{mg/L}$ within hours.

Example 2. Iron oxidation in Pilbara mine tailings. Rio Tinto monitors iron speciation in tailings dams in the Pilbara because $\text{Fe}^{2+}$ released from sulfide ores oxidises to $\text{Fe}^{3+}$, generating acid. The half-equations are $\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$ (oxidation, $+2$ to $+3$) and $\text{O}_2 + 4 \text{H}^+ + 4 e^- \to 2 \text{H}_2 \text{O}$ (reduction, oxygen from $0$ to $-2$). Combined: $4 \text{Fe}^{2+} + \text{O}_2 + 4 \text{H}^+ \to 4 \text{Fe}^{3+} + 2 \text{H}_2 \text{O}$. A standing pond of $10^7 \, \text{L}$ at $\text{Fe}^{2+} = 0.05 \, \text{mol/L}$ holds $500{,}000$ mol of iron(II); fully oxidised, it produces enough $\text{Fe(OH)}_3$ to acidify nearby groundwater by $1$ pH unit. Limestone barriers buffer the discharge.

Try this

Q1. Identify the oxidant, the reductant, and the oxidation state change for each element in: $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$. [3 marks]

• Cue. $\text{Zn}$ reductant ($0 \to +2$, oxidised). $\text{Cu}^{2+}$ oxidant ($+2 \to 0$, reduced).

Q2. Balance the half-equation for permanganate reduction in acidic solution: $\text{MnO}_4^- \to \text{Mn}^{2+}$. State the change in oxidation number. [3 marks]

• Cue. $\text{MnO}_4^- + 8 \text{H}^+ + 5 e^- \to \text{Mn}^{2+} + 4 \text{H}_2 \text{O}$. Mn from $+7$ to $+2$, gained $5$ electrons.

Q3. Iron(II) reacts with permanganate in acidic solution. (a) Write the balanced ionic equation. (b) Calculate moles of $\text{Fe}^{2+}$ oxidised by $25.0 \, \text{mL}$ of $0.0200 \, \text{mol/L}$ $\text{MnO}_4^-$. (c) Explain why the solution colour fades from purple to colourless. [2+2+2 marks]

• Cue. (a) $\text{MnO}_4^- + 5 \text{Fe}^{2+} + 8 \text{H}^+ \to \text{Mn}^{2+} + 5 \text{Fe}^{3+} + 4 \text{H}_2 \text{O}$. (b) $n(\text{MnO}_4^-) = 5.00 \times 10^{-4}$; $n(\text{Fe}^{2+}) = 5 \times 5.00 \times 10^{-4} = 2.50 \times 10^{-3}$ mol. (c) Purple $\text{MnO}_4^-$ reduced to colourless $\text{Mn}^{2+}$.