IB Chemistry HLTopic 6 — Electron TransferPaper 1 & 2Core skill~11 min read
Standard Cell Potentials
Once every half-cell has a number, the rest is arithmetic. Put any two half-cells together, subtract one value from the other, and you get the voltage of the cell — plus a free answer to the bigger question: will this reaction happen at all?
📚 What you need to know
EMF = E(right) − E(left), using the standard cell diagram convention.
Set up so the more negative half-cell is on the left. Then the answer comes out positive and the reaction is spontaneous.
The more positive Eθ, the better that species is at gaining electrons — a stronger oxidising agent.
The more negative Eθ, the better at losing electrons — a stronger reducing agent.
Oxidation happens at the more negative electrode; in a voltaic cell that is the negative terminal.
Never multiply Eθ when you scale a half-equation. Voltage is not a per-mole quantity.
A positive Eθcell says a reaction can go, not that it will go quickly.
The rule, and why it is that way round
A voltmeter measures the potential at its right-hand lead minus the potential at its left-hand lead. That is all EMF = E(right) − E(left) is saying.
Cell potential
Eθcell = Eθ(right) − Eθ(left)
In practice you get to choose which half-cell goes on which side, so choose sensibly: put the more negative one on the left. Subtracting a negative number from a positive one gives a positive answer, and a positive cell potential is the sign of a reaction that runs by itself.
A cleaner way to say the same thing: Eθcell = Eθ(the half-cell being reduced) − Eθ(the half-cell being oxidised). If you can spot which species is grabbing the electrons, you do not need to worry about left and right at all.
The electrochemical series
List every half-cell by its Eθ value and you get a ranking. Anything high on the list will pull electrons off anything lower down.
The gap between two rows is the cell voltage. Fluorine and lithium are 5.91 V apart, which is why that pairing releases so much energy.
Which electrode is which
Careful: this is for voltaic cells. In an electrolytic cell the anode is the positive electrode, because a power supply is forcing the reaction.
The one thing that never changes: oxidation is always at the anode and reduction is always at the cathode. It is only the charge on those electrodes that swaps between voltaic and electrolytic cells.
Reading the numbers
What you see
What it means
Quick example
Large positive Eθ
Strong oxidising agent. Takes electrons off almost anything.
F2, +2.87 V
Small positive Eθ
Mildly oxidising. Will oxidise metals below it only.
Cu2+, +0.34 V
Small negative Eθ
Mildly reducing. Will react with acid to give hydrogen.
Find both half-equations in Section 19 of the booklet, written as reductions.
Spot the more positive one. That one runs forwards as a reduction.
Flip the other. Reverse the more negative half-equation so it becomes an oxidation.
Subtract: Eθcell = more positive − more negative. The answer is always positive.
Balance the electrons to build the overall equation — but leave the voltage alone.
Worked examples
WORKED EXAMPLE
A zinc half-cell (−0.76 V) is connected to a copper half-cell (+0.34 V). Find the cell potential and write the overall equation.
Step 1: Identify the more positive half-cellCopper, +0.34 V. So copper ions are reduced and zinc is oxidised.Step 2: SubtractEθcell = (+0.34) − (−0.76) = +1.10 VStep 3: Write the half-equations the way they runZn(s) → Zn2+(aq) + 2e− Cu2+(aq) + 2e− → Cu(s)Step 4: Add them, cancelling the electronsZn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), Eθcell = +1.10 Vzinc is the negative electrode, so electrons flow zinc to copper
WORKED EXAMPLE
Using Ag+/Ag = +0.80 V and Fe3+/Fe2+ = +0.77 V, will silver ions oxidise iron(II) ions?
Step 1: Both are positive — compare them anywaySilver is the more positive, so silver ions take the electrons.Step 2: SubtractEθcell = 0.80 − 0.77 = +0.03 VStep 3: Positive, so yes — but only justAg+(aq) + Fe2+(aq) → Ag(s) + Fe3+(aq)Yes, spontaneous, Eθcell = +0.03 Va tiny positive value still counts as spontaneous — the sign is what matters
WORKED EXAMPLE
Explain, using Eθ values, why copper does not react with dilute hydrochloric acid.
Step 1: Write the two half-cells involved2H+ + 2e− ⇌ H2 0.00 V Cu2+ + 2e− ⇌ Cu +0.34 VStep 2: For copper to dissolve, it would have to be oxidisedThat makes hydrogen the reduction, so hydrogen is on the right.Eθcell = 0.00 − (+0.34) = −0.34 VNegative, so the reaction is not spontaneous — no reactioncopper is below hydrogen in the reactivity series, and the numbers say the same thing
WORKED EXAMPLE
A copper half-cell (+0.34 V) is placed on the left and an unknown half-cell on the right. The EMF is +1.02 V. Identify the unknown from the series above.
Step 1: Put the numbers into the ruleEMF = E(right) − E(left)+1.02 = E(right) − (+0.34)Step 2: RearrangeE(right) = 1.02 + 0.34 = +1.36 VCl2(g) + 2e− ⇌ 2Cl−(aq), Eθ = +1.36 Vadd the left-hand value back on — do not subtract it twice
💡 Exam tip
Show the subtraction with brackets. Writing 0.34 − (−0.76) makes the double negative obvious and protects your working mark.
Do not multiply Eθ. If you double a half-equation to balance electrons, the voltage stays exactly the same.
Quote the answer with a sign and units: +1.10 V, not 1.10.
“Spontaneous” only needs Eθcell to be positive. It says nothing about speed.
If asked which is the strongest reducing agent, look for the most negative value and name the species on the right of that half-equation.
Standard conditions matter: 1.00 mol dm−3, 100 kPa, 298 K. Outside those, the booklet values are only a guide.
⚠ Common mix-up
Adding instead of subtracting. It is always right minus left. Adding gives you a number that looks plausible and is wrong.
Multiplying the voltage by the number of electrons. Very common, always wrong. Voltage is energy per unit charge, so scaling cancels out.
Naming the wrong species as the oxidising agent. The oxidising agent is the one on the left of the reduction half-equation, the one being reduced.
Assuming a positive EMF means you will see something. Carbon should burn in air at room temperature by this test. It does not, because the activation energy is huge.
Mixing up anode charge. Anode is negative in a voltaic cell, positive in electrolysis. Oxidation happens there either way.
Forgetting to reverse a half-equation before adding. If both are written as reductions, the electrons will not cancel.
Up next: Gibbs Energy and Standard Cell Potential — there is a single equation that links the voltage you have just calculated to the energy the reaction can release.
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