Chemistry · Chapter 17

Electrochemistry

Some reactions can push electrons through a wire. That one fact is why a battery works. In this chapter you will learn to spot those reactions and balance them in halves. Then you will build the cell that puts them to work. You will see why zinc gives up electrons and copper takes them. You will turn a voltage into a free energy and into an equilibrium constant. You will learn why a tired battery reads low. At the end you will run a cell backward on purpose, to plate metal and to keep iron from rusting.

By the end of this chapter you can

Split a redox reaction into two half-reactions, then balance each one in acid or in base.
Sketch a galvanic cell and name the anode, the cathode and the salt bridge.
Trace where the electrons go in the wire and where the ions go in the bridge.
Write any cell in short line notation.
Use a table of standard reduction potentials to get E°_cell for any pair.
Move among E°_cell, ΔG° and K with the right n each time.
Use the Nernst equation to find a cell potential away from standard conditions.
Use Faraday's law to link current, time and the mass of metal plated out.

Section 1Balancing redox half-reactions

Two things happen at once

A redox reaction moves electrons from one species to another. Oxidation is the loss of electrons. Reduction is the gain of electrons. One can never happen alone. The electrons have to come from somewhere and land somewhere.

You track the move with oxidation numbers. An oxidation number is a bookkeeping charge for one atom. If the number goes up, that atom lost electrons. If it goes down, that atom gained them. The reducing agent is the one that gets oxidized. The oxidizing agent is the one that gets reduced. Each name tells you what that species does to its partner.

Split the job in two

Big redox equations fight back if you balance them by eye. So split them. A half-reaction shows one side of the trade, with the electrons written in. Balance each half on its own. Then scale the halves so the electrons cancel.

Here is the recipe in acid.

Write one half-reaction for each atom whose oxidation number changed.
Balance every atom except oxygen and hydrogen.
Add H₂O to the side that needs oxygen.
Add H⁺ to the side that needs hydrogen.
Add electrons until the total charge matches on both sides.
Multiply each half so the two have the same electron count.
Add the halves. Cancel anything that shows up on both sides.

What changes in base

A basic solution holds almost no H⁺. You still balance in acid first, because the steps are easy there. Then add OH⁻ to both sides, one for every H⁺ you used. On the H⁺ side each pair becomes water. Now cancel the water that sits on both sides. Nothing else moves.

Watch that trick on the half-reaction that eats oxygen gas. In acid it reads O₂ + 4H⁺ + 4e⁻ → 2H₂O. Add four OH⁻ to each side and four water molecules appear on the left. Two cancel, and you are left with the basic form. Both forms are correct. You just use the one that fits the beaker.

One recipe, one small add-on. The acid steps build the half-reaction. Adding hydroxide to both sides moves it into base without touching the electron count.

Worked example · purple fades to clear

Permanganate ion oxidizes iron(II) in acid. Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.

Step 1 · two halves

Manganese goes from +7 to +2. Iron goes from +2 to +3.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Fe²⁺ → Fe³⁺ + e⁻

Step 2 · match the electrons

One half drops five electrons. The other gives up one. Multiply the iron half by 5.

5Fe²⁺ → 5Fe³⁺ + 5e⁻

Step 3 · add and cancel

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Step 4 · check it twice

Atoms: 1 Mn, 5 Fe, 4 O and 8 H on each side.

Charge: (−1) + (+10) + (+8) = +17 on the left. (+2) + (+15) = +17 on the right. Balanced.

Key points

Oxidation loses electrons. Reduction gains them. They always travel together.
A half-reaction carries its own electrons, so you can balance it alone.
In acid, add water for oxygen and H⁺ for hydrogen, then electrons for charge.
For base, balance in acid and then add OH⁻ to both sides.
Check atoms and charge every single time. Charge is the one people skip.
The electron count you end up with is the n for every equation later in this chapter.

Interactive Balance half-reactions in acid and in base. Open full ↑

Section 2Galvanic cells

Keep the two halves apart

Drop a strip of zinc into copper(II) nitrate and the trade happens at once. Copper coats the zinc and the solution loses its blue. You feel the beaker warm up. But the electrons jump straight across, so you get no current out of it.

A galvanic cell fixes that. Put each half-reaction in its own beaker. Each beaker holds an electrode, a solid that carries current, sitting in its own solution. One beaker plus its electrode is a half-cell. Now the electrons cannot jump. Their only road is the wire you string between the two metals. Current in a wire is current you can use.

Anode, cathode and the salt bridge

The anode is the electrode where oxidation happens. The cathode is the electrode where reduction happens. Two short phrases lock it in. An ox: anode, oxidation. Red cat: reduction, cathode. Electrons always leave the anode and travel to the cathode.

In a galvanic cell the anode is the negative terminal. It is piling up electrons. The cathode is the positive terminal, pulling them in. Zinc metal dissolves at the anode. Copper metal plates out at the cathode. You can watch the zinc strip get thin and the copper strip get fat.

There is one more problem. The anode beaker gains Zn²⁺ and goes positive. The cathode beaker loses Cu²⁺ and goes negative. Charge piles up and the current stops within seconds. A salt bridge solves it. The bridge is a tube of salty gel that lets ions walk through. Negative ions drift toward the anode. Positive ions drift toward the cathode. Both beakers stay neutral, so the current keeps flowing. A porous disc between the two solutions does the same job.

Writing a cell on one line

Drawing beakers gets old fast. Cell notation packs the same picture into one line. The anode goes on the left and the cathode on the right. A single bar is a phase boundary. A double bar is the salt bridge. Concentrations and pressures go in parentheses.

Zn(s) │ Zn²⁺(1 M) ││ Cu²⁺(1 M) │ Cu(s)

Read it left to right and you are walking the electrons. Zinc gives them up. They cross the wire. Copper ions take them. If a half-cell has no solid of its own, you add an inert electrode such as platinum or graphite. It only ferries electrons and never reacts.

The classic lab cell. Zinc dissolves on the left, copper plates on the right, and the salt bridge keeps both beakers neutral so the current does not stall.

Worked example · read a cell line

A cell is written Mg(s) │ Mg²⁺(1 M) ││ Ag⁺(1 M) │ Ag(s). Name the electrodes and write the overall reaction.

Step 1 · left is the anode

Magnesium sits on the left, so magnesium is the anode. It is oxidized.

Mg → Mg²⁺ + 2e⁻

Step 2 · right is the cathode

Silver sits on the right, so silver ion is reduced on the silver strip.

Ag⁺ + e⁻ → Ag

Step 3 · match electrons and add

Double the silver half so both halves move two electrons.

Mg + 2Ag⁺ → Mg²⁺ + 2Ag

Step 4 · say what you would see

The magnesium strip pits and shrinks. Shiny silver builds up on the other strip. Nitrate ions walk toward the magnesium beaker. Here n is 2.

Key points

A galvanic cell keeps the two half-reactions in separate containers.
Oxidation happens at the anode. Reduction happens at the cathode.
Electrons move through the wire from anode to cathode, always.
In a galvanic cell the anode is negative and the cathode is positive.
The salt bridge passes ions so neither beaker builds up charge.
In cell notation the anode is on the left and the double bar is the bridge.

Interactive Build a cell, name the electrodes and follow the electrons. Open full ↑

Section 3Standard reduction potentials

You can only measure a difference

A voltmeter needs two leads. So there is no way to measure one half-cell by itself. Every reading is a gap between two half-cells. Chemists got around this by picking one half-cell as the zero mark.

That reference is the standard hydrogen electrode, or SHE. It is a platinum strip in 1 M H⁺, with hydrogen gas bubbling over it at 1 bar, at 25 °C. Its potential is defined as exactly 0.000 V. Nothing in nature makes it zero. We just agreed to call it zero, the way sea level is the zero for altitude.

Now pair any half-cell with the SHE and read the meter. That reading is the standard reduction potential, written E°. The little circle means standard conditions: 1 M for every dissolved species, 1 bar for every gas, 25 °C. Every value in the table is written as a reduction, electrons on the left.

Reading the table

A positive E° means that species grabs electrons more eagerly than H⁺ does. A negative E° means it holds them more loosely. Copper sits at +0.34 V, so copper ions pull electrons in. Zinc sits at −0.76 V, so zinc metal would rather let them go. Fluorine tops the table at +2.87 V. Lithium sits at the bottom at −3.04 V.

Two habits follow from that. High on the table means a strong oxidizing agent. Low on the table means a strong reducing agent. And any species high on the table will oxidize anything lower down.

Half-reaction (as a reduction)	E° (V)	What it tells you
F₂ + 2e⁻ → 2F⁻	+2.87	Strongest common oxidizing agent
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51	The purple titration workhorse
Cl₂ + 2e⁻ → 2Cl⁻	+1.36	Bleach chemistry
O₂ + 4H⁺ + 4e⁻ → 2H₂O	+1.23	Air, in acid
Ag⁺ + e⁻ → Ag	+0.80	Silver plates out easily
Fe³⁺ + e⁻ → Fe²⁺	+0.77	No solid metal in this one
O₂ + 2H₂O + 4e⁻ → 4OH⁻	+0.40	Air, in neutral or basic water
Cu²⁺ + 2e⁻ → Cu	+0.34	The usual cathode in the lab
2H⁺ + 2e⁻ → H₂	0.00	The agreed zero, the SHE
Pb²⁺ + 2e⁻ → Pb	−0.13	Barely below hydrogen
Ni²⁺ + 2e⁻ → Ni	−0.25	Plating and coins
Fe²⁺ + 2e⁻ → Fe	−0.44	Why iron rusts
Zn²⁺ + 2e⁻ → Zn	−0.76	The usual anode in the lab
Al³⁺ + 3e⁻ → Al	−1.66	Needs electrolysis to make
Mg²⁺ + 2e⁻ → Mg	−2.37	A sacrificial anode metal
Li⁺ + e⁻ → Li	−3.04	Strongest common reducing agent

Putting two half-cells together

To get the cell potential, take the cathode value minus the anode value.

E°_cell = E°_cathode − E°_anode

Both numbers come straight out of the table as reductions. Do not flip a sign before you subtract. The subtraction already handles the flip. If the answer comes out positive, the cell runs on its own. If it comes out negative, you picked the wrong one to be the anode. Swap them and the sign turns positive.

One rule trips up almost everyone. E° never scales with coefficients. Doubling a half-reaction doubles the electrons and doubles the energy, but the potential stays put. Potential is energy per unit of charge, so the doubling cancels out. It is like density: two bricks of lead are heavier than one, but the density is the same.

One ladder, every pair. Each couple sits at its own height. The vertical gap between two rungs is the standard cell potential, and the higher rung is always the cathode.

Worked example · zinc against silver

Find E°_cell and n for a zinc and silver cell at standard conditions.

Step 1 · pull both values

Ag⁺ + e⁻ → Ag has E° = +0.80 V.

Zn²⁺ + 2e⁻ → Zn has E° = −0.76 V.

Step 2 · higher value is the cathode

Silver is higher, so silver is reduced. Zinc is the anode.

Step 3 · subtract

E°_cell = (+0.80 V) − (−0.76 V) = +1.56 V

The answer is positive, so the cell runs on its own.

Step 4 · find n from the balanced equation

Zn + 2Ag⁺ → Zn²⁺ + 2Ag moves two electrons, so n = 2.

Notice what did not happen. Doubling the silver half did not change +0.80 V to +1.60 V. The potential is per charge, so it holds still.

Key points

No single half-cell potential can be measured. Only gaps can.
The hydrogen electrode is defined as exactly 0.000 V to break the tie.
Every tabulated E° is written as a reduction at 1 M, 1 bar and 25 °C.
E°_cell = E°_cathode − E°_anode, with both values taken straight from the table.
A positive E°_cell means the reaction as written is spontaneous.
Scaling a half-reaction changes n but never changes E°.

Interactive Pick two half-cells and compute the standard cell potential. Open full ↑

Section 4Cell potential, free energy and K

A voltage is energy per charge

One volt means one joule for every coulomb of charge moved. So if you know the charge and the voltage, you know the energy. The charge comes from the electron count. One mole of electrons carries 96,485 coulombs. That number is the Faraday constant, F.

Push n moles of electrons through a cell and the charge is nF. Multiply by the potential and you have the most useful work the cell can do. Free energy change is the mirror of that work, so it carries a minus sign.

ΔG° = −nFE°_cell

Here n is moles of electrons in the balanced overall reaction. It is not a coefficient you can guess. Get it from the half-reactions after you match them. A wrong n wrecks every number that follows.

The signs all agree

Look at the minus sign and the story falls out. A positive E°_cell gives a negative ΔG°. Negative ΔG° means spontaneous. So a cell with a positive potential runs on its own and can light a bulb. A negative potential gives a positive ΔG°, and that reaction needs a push from outside.

If E°_cell is	then ΔG° is	and K is	the reaction
positive	negative	greater than 1	runs forward on its own
zero	zero	equal to 1	sits at equilibrium at standard state
negative	positive	less than 1	needs outside energy to go forward

And K joins the family

From the chapter on free energy you already know ΔG° = −RT ln K. Now you have a second route to the same ΔG°. Set the two equal and the F and R sort themselves out.

−nFE°_cell = −RT ln K → E°_cell = (RT / nF) ln K

At 25 °C the clump RT/F times ln 10 works out to 0.0592 V. Check it yourself. Take 8.314 J/mol·K times 298.15 K, divide by 96,485 C/mol, then multiply by 2.303. That gives the shortcut form in base ten logs.

E°_cell = (0.0592 V / n) log K (at 25 °C)

This is why small voltages hide huge equilibrium constants. Each 0.0592 volts multiplies K by ten, per electron. Two electrons and a volt of potential put you past 10³³. A cell you would call ordinary is a reaction that goes to completion.

One triangle to remember. Cell potential, free energy and the equilibrium constant are three readings of the same reaction. The electron count n is the bridge between them.

Worked example · from 1.10 volts to a giant K

For Zn + Cu²⁺ → Zn²⁺ + Cu, find ΔG° and K at 25 °C.

Step 1 · get E° and n

E°_cell = 0.34 V − (−0.76 V) = 1.10 V. Each half moves two electrons, so n = 2.

Step 2 · free energy

ΔG° = −(2)(96,485 C/mol)(1.10 J/C)

ΔG° = −212,267 J/mol, which rounds to −212 kJ/mol.

Step 3 · equilibrium constant

log K = nE°_cell / 0.0592 V = (2)(1.10) / 0.0592 = 37.162

K = 10^37.162 = 1.5 × 10³⁷

Step 4 · check it a second way

Try ΔG° = −RT ln K. Take ln(1.5 × 10³⁷) = 85.60.

−(8.314)(298.15)(85.60) = −212,000 J/mol. The two routes agree.

Step 5 · say what it means

A K near 10³⁷ is enormous. Put zinc in copper sulfate and the copper ions all but vanish.

Key points

One volt is one joule per coulomb, so voltage times charge gives energy.
F = 96,485 C per mole of electrons.
ΔG° = −nFE°_cell, and n comes from the balanced overall reaction.
Positive E°_cell, negative ΔG° and K above 1 all say the same thing.
At 25 °C, E°_cell = (0.0592 V / n) log K.
Every 0.0592 V per electron is one more power of ten in K.

Interactive Convert between E_cell, ΔG and K. Open full ↑

Section 5The Nernst equation and batteries

Standard conditions almost never hold

Every E° in the table assumes 1 M and 1 bar. Your beaker is rarely that tidy. A battery in use is certainly not. The Nernst equation fixes the potential for whatever concentrations you actually have.

E_cell = E°_cell − (RT / nF) ln Q

At 25 °C you can use the base ten form, which is easier to punch in.

E_cell = E°_cell − (0.0592 V / n) log Q

The Q here is the same reaction quotient you already know. Products over reactants, each raised to its coefficient. Pure solids and pure liquids count as 1. Gases go in as pressures. Build Q from the overall cell reaction, not from one half.

Now read the minus sign. Pile up products and Q rises, log Q goes positive, and the potential drops. Pile up reactants and the potential climbs. That is the whole behavior of a battery in one line. As it runs, products build and voltage sags. When Q finally reaches K, log Q is big enough to cancel E° exactly. The cell reads zero. A dead battery is a cell that reached equilibrium.

A cell made of nothing but a difference

Here is a strange one. Put a copper strip in 0.0010 M Cu²⁺ and another in 1.0 M Cu²⁺, then join them. Both half-cells are the same couple, so E°_cell is exactly zero. Yet the meter reads about 0.089 V.

That is a concentration cell. The only thing driving it is the gap in concentration. Nature wants the two solutions to even out. So copper dissolves on the dilute side and plates on the concentrated side. The dilute half-cell is always the anode. The cell dies when the two concentrations meet. Your nerve cells run on this idea, using a potassium gap instead of a copper one.

Batteries you can buy

A battery is a galvanic cell in a can. An alkaline cell runs zinc against manganese dioxide in potassium hydroxide, near 1.5 V. A lead-acid cell runs lead against lead dioxide in sulfuric acid at about 2.0 V, and six in a row make the 12 V in a car. Lithium-ion cells shuttle Li⁺ between a graphite layer and a metal oxide layer at roughly 3.7 V. A hydrogen fuel cell joins hydrogen with oxygen electrically and stands at 1.23 V standard.

Notice what sets them apart. In a lead-acid cell the reactants and products are mostly solids, so Q barely moves and the voltage holds steady until the acid runs thin. An alkaline cell sags as it drains. A fuel cell never discharges at all, because you keep feeding it fresh fuel.

A straight line with a useful slope. Cell potential drops by 0.0592/n volts for every tenfold rise in Q. Follow the line far enough right and it crosses zero, which is where Q equals K and the cell is finished.

Worked example · a cell that is not standard

A zinc and copper cell has [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M. Find E_cell at 25 °C.

Step 1 · overall reaction and n

Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), with n = 2 and E°_cell = 1.10 V.

Step 2 · write Q

Both metals are pure solids, so they drop out.

Q = [Zn²⁺] / [Cu²⁺] = 1.0 / 0.010 = 100

Step 3 · put it in Nernst

E_cell = 1.10 V − (0.0592 V / 2) log(100)

E_cell = 1.10 V − (0.0296 V)(2) = 1.10 V − 0.0592 V = 1.04 V

Step 4 · does it make sense?

You starved the cell of the reactant Cu²⁺, so the potential should fall. It did, by about 0.06 V. Small change, which is why batteries hold a steady voltage for so long.

Worked example · the concentration cell

Find E_cell for Cu │ Cu²⁺(0.0010 M) ││ Cu²⁺(1.0 M) │ Cu.

Step 1 · standard potential is zero

Both electrodes are copper, so E°_cathode and E°_anode are both +0.34 V. Their difference is 0.

Step 2 · the reaction just moves ions

Cu²⁺(1.0 M) → Cu²⁺(0.0010 M), and n = 2.

Q = 0.0010 / 1.0 = 1.0 × 10⁻³, so log Q = −3.

Step 3 · Nernst does all the work

E_cell = 0 − (0.0592 V / 2)(−3) = +0.0888 V, or 0.089 V.

Step 4 · which side is the anode?

The dilute side. Copper dissolves there to raise its own concentration. The meter stays positive until the two solutions match.

Key points

Nernst corrects E°_cell for real concentrations and pressures.
At 25 °C, E_cell = E°_cell − (0.0592/n) log Q.
Build Q from the overall cell reaction, with solids and pure liquids as 1.
More product means a bigger Q and a smaller potential.
A dead battery is a cell where Q has climbed all the way to K, so E_cell is 0.
A concentration cell has E°_cell = 0, and the dilute half-cell is the anode.

Interactive Change concentrations and watch cell potential move. Open full ↑

Section 6Electrolysis and corrosion

Running a cell backward on purpose

So far the reactions have paid their own way. Now flip it. Electrolysis uses an outside power supply to force a reaction that would never go by itself. The container is an electrolytic cell. Its E°_cell is negative and its ΔG° is positive, so you pay the bill in electricity.

The job names do not change. Oxidation still happens at the anode. Reduction still happens at the cathode. What flips is the sign on the terminals. The anode is now the positive post, because the supply pulls electrons out of it. The cathode is the negative post, because electrons get shoved in.

Melt sodium chloride, run current through it, and you get sodium metal at the cathode and chlorine gas at the anode. That is how sodium is made. It takes roughly four volts and a furnace above 800 °C. The table values are for water at 25 °C, so a molten salt needs its own data.

Salt water behaves differently, because water can react too. The cathode gives hydrogen gas, not sodium metal, since water is far easier to reduce than Na⁺. The anode gives chlorine, and sodium hydroxide builds up in the solution. That one process makes chlorine, lye and hydrogen at once.

Real cells need more voltage than the math says. That extra is called overpotential. It pays for slow steps at the electrode surface, such as gas bubbles forming. Splitting water needs 1.23 V in theory and closer to 1.8 V in practice.

Faraday's law counts the electrons

Current is charge per second, so current times time is charge. Divide the charge by F and you have moles of electrons. That is Faraday's law of electrolysis in plain words. Electrons are the reagent, and you can measure them out with an ammeter and a clock.

moles of e⁻ = It / F then moles of product = moles of e⁻ / n

The n is the electrons per formula unit in the half-reaction. Silver takes one electron per atom, copper two, aluminum three. So the same charge gives three times as many moles of silver as of aluminum.

Corrosion is a cell you did not build

Corrosion is the slow galvanic cell that eats metal in the open air. Iron is the famous victim. One patch of a damp nail acts as the anode and iron dissolves there. Another patch acts as the cathode, where oxygen from the air is reduced.

Fe → Fe²⁺ + 2e⁻ and O₂ + 2H₂O + 4e⁻ → 4OH⁻

Those two give 0.40 V − (−0.44 V) = 0.84 V, so rusting is spontaneous. Air and water are all it needs. The Fe²⁺ then oxidizes further to hydrated iron(III) oxide, which is the flaky orange solid you know as rust. Road salt makes it faster, because the ions carry current between the anode and cathode patches.

You can fight it three ways. Block the water with paint or grease. Or pick a metal that makes a tight oxide skin, the way aluminum does. Rust flakes off and bares fresh iron, but aluminum oxide clings and seals. Or set up a sacrificial anode. Bolt zinc or magnesium to the iron. Both sit lower on the table, so they oxidize instead and the iron is spared. Galvanized steel is that trick as a coating, and it keeps working even when scratched.

Pay or get paid. On the left a power supply forces a reaction uphill. On the right a cheaper metal is wired in so it corrodes in place of the iron, and the iron becomes the protected cathode.

Worked example · plating silver onto a spoon

You run 2.00 A through a silver plating bath for 30.0 minutes. What mass of silver lands on the spoon? Silver is 107.87 g/mol.

Step 1 · charge delivered

Change the time to seconds. 30.0 min × 60 s/min = 1800 s.

q = It = (2.00 C/s)(1800 s) = 3600 C

Step 2 · moles of electrons

3600 C ÷ 96,485 C/mol = 0.03731 mol e⁻

Step 3 · use n from the half-reaction

Ag⁺ + e⁻ → Ag, so n = 1. One electron buys one silver atom.

moles Ag = 0.03731 mol

Step 4 · grams

(0.03731 mol)(107.87 g/mol) = 4.02 g of silver

Step 5 · sanity check

Swap in copper, where n = 2, and the same charge gives half as many moles. That comes to 1.19 g of copper. Fewer grams, and that is what the electron count predicts.

Key points

Electrolysis pays electricity to drive a reaction with a negative E°_cell.
Oxidation is at the anode in every cell. Only the terminal signs swap.
In water, the solvent often reacts instead of the ion you expected.
Overpotential is the extra voltage a real cell needs above the calculated value.
Moles of electrons = It/F, then divide by n for moles of product.
Rusting is a galvanic cell worth 0.84 V, and a zinc or magnesium anode stops it.

Interactive Drive a nonspontaneous reaction and protect iron from rust. Open full ↑

The chapter in one breath

Split, then balance. Two half-reactions, water for oxygen, H⁺ for hydrogen, electrons for charge. Add OH⁻ to both sides to move into base.
Keep the halves apart. A galvanic cell forces the electrons through a wire instead of letting them jump.
An ox, red cat. Oxidation at the anode, reduction at the cathode, in every cell ever built.
The bridge keeps the peace. Ions move through the salt bridge so neither beaker builds up charge.
Zero is a choice. The hydrogen electrode was set to 0.000 V so every other half-cell could be measured against it.
Subtract, do not flip. E°_cell = E°_cathode − E°_anode, and E° never scales.
One fact, three faces. ΔG° = −nFE°_cell and E°_cell = (0.0592/n) log K at 25 °C.
Nernst explains a dying battery. As product builds, Q rises and the voltage falls. At Q = K the cell reads zero.
Electrons are a reagent. It/F gives moles of electrons, and that is how much metal you can plate or how much rust you can prevent.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back here whenever a term stops making sense.

Anode: The electrode where oxidation happens. Negative in a galvanic cell, positive in an electrolytic one.
Cathode: The electrode where reduction happens. Positive in a galvanic cell, negative in an electrolytic one.
Cell notation: A one-line shorthand for a cell. Anode on the left, single bar for a phase boundary, double bar for the salt bridge.
Concentration cell: A cell whose two half-cells use the same couple at different concentrations. Its E°_cell is zero, so all of its voltage comes from the Nernst term.
Corrosion: Unwanted oxidation of a metal by its surroundings. Rusting iron is a galvanic cell with air as the oxidizing agent.
Electrolysis: Using an outside power supply to drive a reaction that has a negative cell potential. The container it happens in is an electrolytic cell.
Faraday constant: The charge on one mole of electrons, 96,485 C/mol, written F.
Faraday's law of electrolysis: The amount of substance changed at an electrode is set by the charge passed. Moles of electrons equal It/F.
Galvanic cell: A cell that uses a spontaneous redox reaction to push current through an outside circuit. Also called a voltaic cell.
Half-reaction: One side of a redox trade, written with its own electrons so it can be balanced alone.
Nernst equation: The rule that corrects cell potential for real conditions. E_cell = E°_cell − (0.0592/n) log Q at 25 °C.
Overpotential: Extra voltage a real cell needs beyond the calculated minimum, spent on slow steps at the electrode surface.
Oxidation: Loss of electrons, which raises the oxidation number.
Oxidizing agent: The species that takes the electrons and is itself reduced.
Reducing agent: The species that hands over the electrons and is itself oxidized.
Reduction: Gain of electrons, which lowers the oxidation number.
Sacrificial anode: A block or coating of a metal with a lower E° that is wired to a structure so it corrodes in its place.
Salt bridge: A path that lets ions move between the two half-cells so neither one builds up charge.
Standard hydrogen electrode: Platinum in 1 M H⁺ with H₂ at 1 bar and 25 °C, defined as exactly 0.000 V.
Standard reduction potential: The potential of a half-reaction written as a reduction, measured against the hydrogen electrode at standard conditions.

Check yourself

1. In Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), which species is the reducing agent, and which is reduced?

Zinc metal is the reducing agent. Cu²⁺ is reduced. Zinc goes from 0 to +2, so it lost electrons and was oxidized. The thing that gets oxidized is the reducing agent. Copper goes from +2 to 0, so it gained them.

2. Balance this in acid: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺. What is n?

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O, and n = 6. The halves are Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and Fe²⁺ → Fe³⁺ + e⁻, so the iron half is multiplied by 6. Charge check: (−2) + (+12) + (+14) = +24 on the left, and (+6) + (+18) = +24 on the right.

3. Write the cell notation for a cell made from an iron electrode in 1 M Fe²⁺ and a silver electrode in 1 M Ag⁺.

Fe(s) │ Fe²⁺(1 M) ││ Ag⁺(1 M) │ Ag(s) Iron sits at −0.44 V and silver at +0.80 V. Silver is higher, so silver is the cathode and goes on the right. Iron is the anode and goes on the left.

4. Find E°_cell for Mg(s) │ Mg²⁺ ││ Fe²⁺ │ Fe(s), and give n.

+1.93 V, with n = 2. The cathode is iron at −0.44 V and the anode is magnesium at −2.37 V. So E°_cell = (−0.44 V) − (−2.37 V) = +1.93 V. The overall reaction Mg + Fe²⁺ → Mg²⁺ + Fe moves two electrons.

5. For that magnesium and iron cell, what is ΔG°?

−372 kJ/mol ΔG° = −nFE°_cell = −(2)(96,485 C/mol)(1.93 J/C) = −372,432 J/mol. That rounds to −372 kJ/mol. The sign is negative, which matches the positive cell potential.

6. Find K at 25 °C for Cu(s) + 2Ag⁺ → Cu²⁺ + 2Ag(s). Watch the value of n.

K = 3.5 × 10¹⁵ First E°_cell = (+0.80 V) − (+0.34 V) = +0.46 V. The silver half had to be doubled to match copper's two electrons, so n = 2, not 1. Then log K = nE°/0.0592 = (2)(0.46)/0.0592 = 15.54, and K = 10^15.54 = 3.5 × 10¹⁵.

7. A zinc and copper cell has [Zn²⁺] = 0.010 M and [Cu²⁺] = 1.0 M. What does the meter read?

1.16 V Q = [Zn²⁺]/[Cu²⁺] = 0.010/1.0 = 0.010, so log Q = −2. Then E_cell = 1.10 V − (0.0592/2)(−2) = 1.10 V + 0.0592 V = 1.16 V. Less product means a higher potential than standard.

8. Find E_cell for Ag │ Ag⁺(0.010 M) ││ Ag⁺(1.0 M) │ Ag.

+0.118 V Both sides are the same couple, so E°_cell = 0. Here n = 1, and Q = 0.010/1.0 = 0.010, so log Q = −2. Then E_cell = 0 − (0.0592/1)(−2) = +0.118 V. The dilute side is the anode.

9. You electrolyze salt water. What comes off each electrode, and what builds up in the solution?

Hydrogen gas at the cathode, chlorine gas at the anode, and sodium hydroxide in the solution. Water is much easier to reduce than Na⁺, so you never get sodium metal from a water solution. Chloride is oxidized at the anode. The leftover Na⁺ and OH⁻ make lye.

10. How long must you run 5.00 A to plate 10.0 g of copper? Copper is 63.55 g/mol.

About 6.07 × 10³ s, which is 1.69 hours. First, 10.0 g ÷ 63.55 g/mol = 0.1574 mol Cu. Since Cu²⁺ + 2e⁻ → Cu, that needs 0.3147 mol of electrons. Charge: (0.3147 mol)(96,485 C/mol) = 30,365 C. Time: 30,365 C ÷ 5.00 C/s = 6073 s, or 101 minutes.

11. Run 10.0 A for exactly 1.00 hour through molten aluminum oxide. How much aluminum do you get? Aluminum is 26.98 g/mol.

3.36 g Charge: (10.0 C/s)(3600 s) = 36,000 C. Moles of electrons: 36,000 ÷ 96,485 = 0.3731 mol. Since Al³⁺ + 3e⁻ → Al, divide by 3 to get 0.1244 mol Al. Mass: (0.1244)(26.98) = 3.36 g. Three electrons per atom is why aluminum is so costly to make.

12. A galvanized bucket gets a deep scratch down to bare iron. Why does the iron still not rust?

The zinc coating acts as a sacrificial anode. Zinc sits at −0.76 V and iron at −0.44 V, so zinc gives up electrons more readily. The zinc corrodes and feeds electrons to the iron, which makes the iron the cathode. A cathode gains electrons instead of dissolving, so it survives.

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