Honors Chemistry Unit 11 Review: Redox Reactions & Electrochemistry

Redox is short for reduction-oxidation, and the two always come as a pair. Oxidation is losing electrons and reduction is gaining them. The mnemonic OIL RIG (Oxidation Is Losing, Reduction Is Gaining) covers the definitions.

You track electrons with oxidation numbers. Pure elements are 0, monatomic ions equal their charge, oxygen is usually −2 (−1 in peroxides like $\mathrm{H_2O_2}$), hydrogen is usually +1 (−1 in metal hydrides like NaH), and fluorine is −1 in its compounds (elemental $\mathrm{F_2}$ is 0, like any pure element). All the oxidation numbers add up to the overall charge. If an atom's number goes up, it was oxidized. If it goes down, it was reduced.

Worked example. Find the oxidation number of S in $\mathrm{H_2SO_4}$. The compound is neutral, so $2(+1) + S + 4(-2) = 0$, which gives $S = +6$.

Now look at $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$. Zinc goes from 0 to +2, so it is oxidized and acts as the reducing agent. Copper goes from +2 to 0, so it is reduced and acts as the oxidizing agent.

The agent names feel backwards at first. The oxidizing agent causes oxidation in something else by taking its electrons, so the oxidizing agent itself ends up reduced.

A balanced redox equation conserves atoms and charge, which means electrons lost must equal electrons gained. The half-reaction method handles this step by step.

In acidic solution, split the reaction into an oxidation half and a reduction half. Balance atoms other than O and H, add $\mathrm{H_2O}$ to balance oxygen, add $\mathrm{H^+}$ to balance hydrogen, then add electrons to the more positive side to balance charge. Multiply so both halves transfer the same number of electrons, add them, and cancel anything on both sides. For basic solution, finish the acidic version, add one $\mathrm{OH^-}$ to both sides for every $\mathrm{H^+}$, combine $\mathrm{H^+}$ and $\mathrm{OH^-}$ into water, and cancel extra water.

Worked example. Balance $\mathrm{MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}}$ in acid. In the reduction half, Mn drops from +7 to +2. Add 4 $\mathrm{H_2O}$ on the right for oxygen and 8 $\mathrm{H^+}$ on the left for hydrogen. The left charge is $-1 + 8 = +7$ and the right is +2, so add 5 electrons to the left. This gives $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$. The oxidation half, $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$, gets multiplied by 5. Adding them gives:

$\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$

Check the charge. The left side is $-1 + 8 + 10 = +17$ and the right side is $2 + 15 = +17$.

A galvanic (voltaic) cell runs a spontaneous redox reaction in two separate half-cells, so electrons have to travel through a wire. Oxidation happens at the anode and reduction at the cathode (An Ox, Red Cat). Electrons flow from anode to cathode through the wire. A salt bridge lets ions drift between the half-cells so charge doesn't build up and stop the reaction.

Standard reduction potentials (E°) rank reduction half-reactions under standard-state conditions. Dissolved substances have an activity of 1, often approximated as 1 M, gases are at 1 bar, and pure solids and liquids are in their standard forms. Tables are commonly reported at 25°C, with the standard hydrogen electrode assigned 0.00 V. A more positive E° means a stronger oxidizing agent. A negative E° doesn't mean oxidation happens by itself. It only means the species pulls electrons more weakly than H⁺.

Worked example. Find E°cell for a zinc-copper cell. The table gives Cu²⁺/Cu = +0.34 V and Zn²⁺/Zn = −0.76 V. Copper has the higher value, so it is the cathode and zinc is the anode. $E^\circ_{cell} = 0.34 - (-0.76) = +1.10$ V. The positive value means the reaction is spontaneous under standard-state conditions.

An electrolytic cell uses an outside power source to force a non-spontaneous reaction, such as splitting water. Oxidation is still at the anode, but the electrode signs flip.

A battery operates as a galvanic cell while it discharges. A primary battery, like a standard alkaline AA, is not designed for recharging because the chemical and physical changes during discharge are not reliably reversible. A secondary battery, like the lithium-ion cell in your phone, can be recharged because an outside power source pushes electrons backward and rebuilds the reactants. While charging, the battery acts as an electrolytic cell.

A hydrogen fuel cell keeps feeding in its reactants. $\mathrm{H_2}$ is oxidized at the anode, $\mathrm{O_2}$ is reduced at the cathode, and the overall reaction $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ produces electricity with water as the only product.

Corrosion is unwanted redox. Iron is oxidized by oxygen and water to form rust. Galvanizing coats iron with zinc, which has a lower E° (−0.76 V vs. −0.44 V for iron), so the zinc oxidizes first and protects the iron. This is called a sacrificial anode.

Electroplating uses an electrolytic cell to reduce metal ions onto an object at the cathode. Some honors courses also calculate how much metal plates out.

Worked example. A current of 2.00 A runs for 965 s through a $\mathrm{Cu^{2+}}$ solution. Charge is $2.00 \times 965 = 1930$ C. Moles of electrons are $1930 \div 96{,}485 = 0.0200$ mol. Each $\mathrm{Cu^{2+}}$ needs 2 electrons, so 0.0100 mol Cu plates out, which is $0.0100 \times 63.55 = 0.636$ g.