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Standard electrode potentials: WACE Year 12 Chemistry

Syllabus dot point

“Use the standard electrode potential series to predict the spontaneity of redox reactions and calculate standard cell potentials”

8 min read

Quick answer

The standard electrode potential series ranks half-reactions by their tendency to be reduced, all measured against the standard hydrogen electrode (defined as 0 volts). The more positive the potential, the stronger the oxidising agent. A reaction is spontaneous if the standard cell potential Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} is positive.

What this dot point is asking

The standard electrode potential (E∘E^\circ) of a half-reaction measures its tendency to gain electrons (to be reduced) under standard conditions, relative to a reference. Because we cannot measure a single electrode in isolation, all values are quoted against the standard hydrogen electrode, which is assigned a potential of exactly 0 volts.

Standard electrode potential

The voltage of a half-cell under standard conditions (298 K, 100 kPa, 1 mol L−1^{-1} solutions) measured relative to the standard hydrogen electrode, which is defined as 0 V. Values are tabulated as reduction potentials.

Reading the series

In the SCSA data booklet the half-reactions are written as reductions and listed by E∘E^\circ. The key interpretations:

  • A more positive E∘E^\circ means the species on the left is a stronger oxidising agent (it gains electrons readily). Fluorine and permanganate sit near the top.
  • A more negative E∘E^\circ means the species on the right is a stronger reducing agent (it loses electrons readily). Lithium and potassium sit near the bottom.
Stronger oxidiser is higher in the table

The half-reaction with the more positive potential proceeds as written (reduction, at the cathode). The other half-reaction is forced to run in reverse (oxidation, at the anode). The strongest oxidising agents are at the top of the series and the strongest reducing agents at the bottom.

Calculating standard cell potential

For a complete cell, identify which half-reaction is reduction (cathode) and which is oxidation (anode), then:

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

where both values are taken as the tabulated reduction potentials. A positive Ecell∘E^\circ_{cell} means the reaction is spontaneous (it occurs in a galvanic cell). A negative value means it is non-spontaneous and would need an external power source (electrolysis).

Importantly, E∘E^\circ is an intensive property: it does not depend on the amount of substance, so you never multiply a potential by the coefficients used to balance electrons.

Predicting spontaneity

To predict whether a given oxidising agent will react with a given reducing agent, find both half-reactions in the table. If the oxidising agent's half-reaction is higher (more positive) than the reducing agent's, the reaction is spontaneous; the combined Ecell∘E^\circ_{cell} will be positive.

Calculating a cell potential

Question

A cell is built from a zinc electrode and a copper electrode. Given E∘(Cu2+/Cu)=+0.34E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34 V and E∘(Zn2+/Zn)=−0.76E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 V, find the standard cell potential and state whether the cell reaction is spontaneous.

Solution

The copper half-reaction has the more positive potential, so copper ions are reduced (cathode) and zinc is oxidised (anode).

Ecell∘=Ecathode∘−Eanode∘=(+0.34)−(−0.76)=+1.10 VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (+0.34) - (-0.76) = +1.10 \text{ V}

Because Ecell∘E^\circ_{cell} is positive, the reaction Zn+Cu2+→Zn2++Cu\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} is spontaneous, which is why this combination forms a working galvanic cell.

Limitations: thermodynamics, not kinetics

A crucial caveat the best answers include is that E∘E^\circ predicts only whether a reaction is thermodynamically feasible, not whether it will actually proceed at a useful rate. A positive Ecell∘E^\circ_{\text{cell}} says the reaction can happen, but a high activation energy may make it immeasurably slow (hydrogen and oxygen have a very positive cell potential yet do not react without a spark or catalyst). Standard potentials also apply strictly to standard conditions; changing concentration, temperature or using non-standard pressures shifts the real cell voltage away from Ecell∘E^\circ_{\text{cell}}. In aqueous electrolysis, an additional overpotential can mean the predicted product is not the one observed, which is why concentrated chloride solutions liberate chlorine rather than oxygen at the anode.

Why this matters

The potential series is the predictive engine of electrochemistry. It tells you which metal will displace another, which way electrons flow in a cell, what voltage a cell delivers, and which products form during electrolysis. These limitations and predictions appear throughout the redox section of the examination.

Multiplying potentials or mixing up the subtraction

The most common error is multiplying an electrode potential by the number of electrons when balancing; E∘E^\circ is intensive and is never scaled. Another is getting the subtraction backwards or subtracting the wrong sign, especially with a negative anode value. Always identify cathode and anode first, then apply Ecathode∘−Eanode∘E^\circ_{cathode} - E^\circ_{anode} carefully with signs.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks
Use the standard reduction potentials MnO4−/Mn2+=+1.51 V\text{MnO}_4^-/\text{Mn}^{2+} = +1.51\ \text{V}, Fe3+/Fe2+=+0.77 V\text{Fe}^{3+}/\text{Fe}^{2+} = +0.77\ \text{V} and I2/I−=+0.54 V\text{I}_2/\text{I}^- = +0.54\ \text{V}. (a) Predict whether acidified MnO4−\text{MnO}_4^- will oxidise Fe2+\text{Fe}^{2+}, justifying with Ecell∘E^\circ_{\text{cell}}. (b) Predict whether Fe3+\text{Fe}^{3+} will oxidise I−\text{I}^-, justifying with Ecell∘E^\circ_{\text{cell}}. (c) State the strongest oxidising agent listed.
Show worked answer →

A 6 mark question rewards two spontaneity predictions with calculations and the agent identification.

(a) MnO4−\text{MnO}_4^- (cathode, +1.51+1.51) versus Fe2+\text{Fe}^{2+} oxidised (anode, Fe3+/Fe2+\text{Fe}^{3+}/\text{Fe}^{2+} at +0.77+0.77):

Ecell∘=1.51−0.77=+0.74 V.E^\circ_{\text{cell}} = 1.51 - 0.77 = +0.74\ \text{V}.

Positive, so the reaction is spontaneous: permanganate will oxidise Fe2+\text{Fe}^{2+}.

(b) Fe3+\text{Fe}^{3+} reduced (cathode, +0.77+0.77) versus I−\text{I}^- oxidised (anode, I2/I−\text{I}_2/\text{I}^- at +0.54+0.54):

Ecell∘=0.77−0.54=+0.23 V.E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23\ \text{V}.

Positive, so Fe3+\text{Fe}^{3+} will oxidise I−\text{I}^- to iodine.

(c) The species with the most positive reduction potential is the strongest oxidising agent, so MnO4−\text{MnO}_4^- (in acid, +1.51 V+1.51\ \text{V}).

Markers reward both Ecell∘E^\circ_{\text{cell}} values with correct signs and conclusions, and MnO4−\text{MnO}_4^- as the strongest oxidiser.

WACE 2023-style4 marks
Explain why a standard electrode potential cannot be measured for a single half-cell in isolation, and describe how the standard hydrogen electrode solves this problem. Why is E∘E^\circ never multiplied when balancing a half-equation?
Show worked answer →

A 4 mark answer needs the reference-electrode idea and the intensive-property point.

A potential difference can only be measured between two electrodes, so a single half-cell has no absolute potential that can be read directly. To create a common scale, the standard hydrogen electrode (with H+\text{H}^+ at 1 mol L−11\ \text{mol L}^{-1}, H2\text{H}_2 at 100 kPa100\ \text{kPa}, 298 K298\ \text{K}) is assigned a potential of exactly 0 V0\ \text{V}. Every other half-cell is measured against it, giving each a relative standard electrode potential.

E∘E^\circ is an intensive property: it measures the tendency per electron transferred and does not depend on the amount of substance. Scaling a half-equation changes the number of moles reacting but not this per-electron tendency, so E∘E^\circ is never multiplied.

Markers reward the two-electrode measurement point, the hydrogen electrode as the 0 V0\ \text{V} reference, and the intensive-property justification.

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Sources & how we know this

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