Redox reactions: oxidation numbers, balancing by half reactions, a galvanic cell's voltage and what an electrolytic cell deposits

A redox reaction moves electrons: one species is oxidized (its oxidation number rises) and another is reduced (its oxidation number falls), and the electrons lost must equal the electrons gained. That one rule balances the equation, gives a galvanic cell its voltage and tells an electrolytic cell how much metal a current deposits. Below, each is worked on one reaction with every line shown, the wrong line named, and a sibling left for you.

Oxidation numbers by rule

An oxidation number is a bookkeeping charge: the charge an atom would carry if every bond were ionic. The rules, applied in order of priority, with an earlier rule winning when two disagree:

1. An element on its own is 0: Fe(s), O₂(g), H₂(g).
2. A monatomic ion is its charge: Na⁺ is +1, Cl⁻ is −1.
3. Fluorine is −1 in every compound.
4. Hydrogen is +1, except with a metal, where it is −1.
5. Oxygen is −2, except in peroxides, where it is −1.
6. The oxidation numbers in a species add up to its overall charge: zero for a neutral molecule, the ion's charge for an ion.

Rule 6 is the one you solve with. Three worked:

• Mn in MnO₄⁻: x + 4 × (−2) = −1, so x − 8 = −1, so x = +7
• Cr in Cr₂O₇²⁻: 2x + 7 × (−2) = −2, so 2x − 14 = −2, so 2x = 12, so x = +6 (per chromium)
• S in SO₄²⁻: x + 4 × (−2) = −2, so x = +6

Oxidation is a rise in oxidation number; reduction is a fall. Fe²⁺ → Fe³⁺ is oxidation, +2 to +3. Mn in MnO₄⁻ going to Mn²⁺ is reduction, +7 to +2.

The common wrong line. Oxygen as −2 in hydrogen peroxide, H₂O₂, where it is −1 (two hydrogens at +1 and two oxygens must sum to zero). The second: forgetting the ion's overall charge in rule 6 and setting the sum to zero for MnO₄⁻, which gives Mn as +8.

In SCH4U this is three ideas of strand F: Oxidation numbers and balancing redox reactions, Galvanic cells and standard reduction potentials and Electrolytic cells and Faraday's laws; the SCH4U course map places them.

Oxidizing agent, reducing agent: who gets what

The names describe what a species does to the other one. The oxidizing agent oxidizes something else, and to do that it takes electrons, so it is itself reduced. The reducing agent gives electrons away, so it is itself oxidized.

In the reaction between permanganate and iron(II), MnO₄⁻ takes electrons and its manganese falls from +7 to +2: permanganate is the oxidizing agent. Fe²⁺ gives an electron and rises to Fe³⁺: iron(II) is the reducing agent.

The common wrong line. Naming the species that is oxidized as the oxidizing agent. The species that is oxidized is the reducing agent; it gave the electrons.

Worked example: balancing in acid by half reactions

The problem. Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

The half reaction method splits the reaction into the oxidation and the reduction, balances each on its own, and recombines them so that the electrons cancel.

• Oxidation half: Fe²⁺ → Fe³⁺ + e⁻ (one electron, +2 to +3; atoms already balanced)
• Reduction half, atoms other than O and H first: MnO₄⁻ → Mn²⁺ (one Mn each side)
• Balance O with water: MnO₄⁻ → Mn²⁺ + 4 H₂O (four oxygens need four waters)
• Balance H with H⁺: MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O (eight hydrogens in the water need eight H⁺; this is the acidic-solution step)
• Balance charge with electrons: left side −1 + 8 = +7; right side +2; add 5 e⁻ to the left: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O (now +2 each side; Mn going +7 to +2 is a gain of 5, which agrees)
• Equalize the electrons: the iron half gives 1 and the manganese half takes 5, so multiply the iron half by 5: 5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻
• Add, and cancel the 5 e⁻ on each side: MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O

Check atoms. Mn 1 = 1; O 4 = 4; H 8 = 8; Fe 5 = 5.

Check charge. Left: −1 + 5 × (+2) + 8 × (+1) = −1 + 10 + 8 = +17. Right: +2 + 5 × (+3) = +2 + 15 = +17. Both checks pass, and both are required: an equation can balance its atoms and still be wrong on charge.

The common wrong line. Adding the half reactions without equalizing the electrons: MnO₄⁻ + Fe²⁺ + 8 H⁺ → Mn²⁺ + Fe³⁺ + 4 H₂O. Its atoms balance. Its charge does not: left −1 + 2 + 8 = +9, right +2 + 3 = +5. The charge check catches it, which is why the check is not optional. The second wrong line: balancing H with H₂ or O with O₂. In solution the hydrogen and oxygen come from water and H⁺, not from gases.

In basic solution, balance as in acid, then add one OH⁻ to each side for every H⁺, combine H⁺ and OH⁻ into water, and cancel the water that appears on both sides.

A galvanic cell's voltage from the table

A galvanic cell separates the two half reactions into two half-cells joined by a wire and a salt bridge, so the electrons travel through the wire and can do work. The parts and their rules: oxidation happens at the anode, which is the negative electrode; reduction happens at the cathode, which is positive; electrons flow through the wire from anode to cathode; the salt bridge lets ions move so neither half-cell builds up charge.

The table of standard reduction potentials lists every half reaction written as a reduction, with its E° in volts. The rule for a cell:

E°cell = E°(cathode) − E°(anode)

Both potentials are read from the table as reduction potentials, as printed. Neither is multiplied by a coefficient, because a potential is energy per electron and does not scale with how many electrons the equation shows.

The problem. Find the voltage of the zinc–copper cell, given the table values Cu²⁺ + 2 e⁻ → Cu, E° = +0.34 V, and Zn²⁺ + 2 e⁻ → Zn, E° = −0.76 V.

• The higher reduction potential is the one that is reduced: copper is the cathode, zinc the anode
• E°cell = 0.34 − (−0.76) = 0.34 + 0.76 = +1.10 V (cathode minus anode; subtracting a negative)
• Positive, so the cell runs spontaneously as written: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

The common wrong line. Flipping the sign of the anode's potential because it is being oxidized, and then also subtracting: 0.34 − 0.76 = −0.42 V, which would say the cell does not run. Do one or the other, never both; the rule above has the subtraction do the flipping. The second: doubling a potential because a half reaction was doubled when the electrons were equalized. Potentials are per electron already.

An electrolytic cell and Faraday's law

An electrolytic cell is the galvanic cell run backwards: a power supply pushes electrons the way they would not go on their own, driving the non-spontaneous direction. Oxidation is still at the anode and reduction still at the cathode, but now the anode is the positive terminal, because the supply is pulling electrons out of it.

Faraday's law is the arithmetic of how much a current deposits, in three lines:

• Charge: Q = I × t, in coulombs, with t in seconds
• Moles of electrons: n(e⁻) = Q ÷ F, with F = 96,485 C/mol (the charge on one mole of electrons, given here)
• Moles of product: n(e⁻) ÷ (electrons per ion in the half reaction), then × molar mass for grams

The problem. A current of 2.00 A flows for 30.0 min through a copper(II) sulfate solution. What mass of copper is deposited at the cathode? Copper's molar mass is 63.55 g/mol.

• t = 30.0 min × 60 s/min = 1800 s (seconds, always)
• Q = 2.00 A × 1800 s = 3600 C (an ampere is a coulomb per second)
• n(e⁻) = 3600 C ÷ 96,485 C/mol = 0.03731 mol (dividing by F)
• Cu²⁺ + 2 e⁻ → Cu, so n(Cu) = 0.03731 ÷ 2 = 0.01866 mol (two electrons per copper atom)
• m = 0.01866 mol × 63.55 g/mol = 1.186 g → 1.19 g (three significant figures, from the 2.00 and the 30.0)

The moles here are the stoichiometry page's moles once more, with the electron as one more species in the ratio.

The common wrong line. Minutes left as minutes: Q = 2.00 × 30.0 = 60 C, and every later line a thirtieth of the right size. The second: forgetting the 2 electrons per copper ion, which gives 2.37 g, double the answer.

Redox on one screen

Piece	The rule
Oxidation numbers	Element 0; ion its charge; F −1; H +1 (−1 with metals); O −2 (−1 in peroxides); the sum is the species' charge
Half reactions in acid	Other atoms; O with H₂O; H with H⁺; charge with e⁻; equalize; add; check atoms and charge
Agents	The oxidizing agent is reduced; the reducing agent is oxidized
Cell voltage	E°cell = E°(cathode) − E°(anode), both as reduction potentials, never multiplied
Faraday	Q = It (seconds); n(e⁻) = Q ÷ 96,485; divide by electrons per ion; × molar mass

Attempting the problems below before reading their answers is chapter 3 of the free playbook.

Try one cold

Cover the page above. Three problems.

(a) Balance Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ in acidic solution, and check the charge.

(b) A cell is built from Ag⁺ + e⁻ → Ag, E° = +0.80 V, and Ni²⁺ + 2 e⁻ → Ni, E° = −0.26 V (the table values given). Which is the cathode, and what is E°cell?

(c) The same 2.00 A for 30.0 min passes through a silver nitrate solution. What mass of silver is deposited? Ag⁺ + e⁻ → Ag; silver's molar mass is 107.87 g/mol.

Check

• (a) Reduction half: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O. Seven oxygens take seven waters, whose fourteen hydrogens take fourteen H⁺; the charge on the left is then −2 + 14 = +12 against +6 on the right, so six electrons; each chromium falls +6 to +3, two of them, six electrons, which agrees. Oxidation half × 6: 6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻. Overall: Cr₂O₇²⁻ + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O. Charge check: left −2 + 12 + 14 = +24; right +6 + 18 = +24.
• (b) Silver has the higher reduction potential, so silver is the cathode and nickel the anode. E°cell = 0.80 − (−0.26) = +1.06 V.
• (c) Q = 3600 C and n(e⁻) = 0.03731 mol as before. One electron per silver ion, so n(Ag) = 0.03731 mol. m = 0.03731 × 107.87 = 4.02 g.

If (a) failed its charge check, look at whether the electrons were equalized. If (b) came out as 0.54 V, you flipped and subtracted. If (c) came out as 2.01 g, you divided by two electrons that silver does not need.

Where this sits in the course

These are ideas 21, 22 and 23 of 24 in SCH4U, Oxidation numbers and balancing redox reactions, Galvanic cells and standard reduction potentials and Electrolytic cells and Faraday's laws: the whole of strand F, the last strand of the course. They follow strand E's Buffers and acid–base titration, and the H⁺ and water used to balance in acid are the acid–base idea in a new place.

They rest on two earlier ideas. Strand C's Electron configurations and periodic trends explains why metals tend to give up electrons, the tendency a low reduction potential measures; the electron configuration page has the trend. And the last idea in the pack, Gas laws and stoichiometry (prerequisite review), supplies the moles in Faraday's law.

For US readers, the AP Chemistry exam covers the same three pieces. Every idea of the three science courses has a page like this one.

How to practise it

Mix the four kinds in one set: oxidation numbers, a balance in acid, a cell voltage, an electrolysis mass. Rohrer and Taylor (2007) found that students who practised problem types shuffled together did worse during practice and much better on a test a week later than students who practised each type in its own block, because the mixed set makes you choose the method.

Run the charge check on every balanced equation before you look at the answer. Roediger and Karpicke (2006) found that testing yourself on material beats restudying it for recall a week later, and a charge check is a test you set yourself, with the answer hidden until you have committed. Active recall has the honest-marking rules, and how to study chemistry treats this strand as four procedures, each a short list of lines.