SCSA WACE 2021-style practice: Use the standard reduction potentials MnO_4^-/Mn^2+ = +1.51\ V, Fe^3+/Fe^2+ = +0.77\ V and I_2/I^- = +0.54\ V. (a) Predict whether acidified MnO_4^- will oxidise Fe^2+, justifying with E^_cell. (b) Predict whether Fe^3+ will oxidise I^-, justifying with E^_cell. (c) State the strongest oxidising agent listed.

A 6 mark question rewards two spontaneity predictions with calculations and the agent identification. **(a)** MnO_4^- (cathode, +1.51) versus Fe^2+ oxidised (anode, Fe^3+/Fe^2+ at +0.77): $E^_cell = 1.51 - 0.77 = +0.74\ V.$ Positive, so the reaction is spontaneous: permanganate will oxidise Fe^2+. **(b)** Fe^3+ reduced (cathode, +0.77) versus I^- oxidised (anode, I_2/I^- at +0.54): $E^_cell = 0.77 - 0.54 = +0.23\ V.$ Positive, so Fe^3+ will oxidise I^- to iodine. **(c)** The species with the most positive reduction potential is the strongest oxidising agent, so MnO_4^- (in acid, +1.51\ V). Markers reward both E^_cell values with correct signs and conclusions, and MnO_4^- as the strongest oxidiser.

§-Syllabus dot point

WAChemistrySyllabus dot point

How does the standard electrode potential series let us predict whether a redox reaction will occur and how much voltage it produces?

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

A focused answer to the WACE Year 12 Chemistry dot point on standard electrode potentials, the reference hydrogen electrode, predicting spontaneity, and calculating standard cell EMF from the potential series, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

The standard electrode potential ( $E^\circ$ ) 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.

Reading the series

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

A more positive $E^\circ$ 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 $E^\circ$ means the species on the right is a stronger reducing agent (it loses electrons readily). Lithium and potassium sit near the bottom.

Calculating standard cell potential

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

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

where both values are taken as the tabulated reduction potentials. A positive $E^\circ_{cell}$ 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, $E^\circ$ 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 $E^\circ_{cell}$ will be positive.

Calculating a cell potential

Question

A cell is built from a zinc electrode and a copper electrode. Given $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V and $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76$ 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).

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (+0.34) - (-0.76) = +1.10 \text{ V}

Because $E^\circ_{cell}$ is positive, the reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ 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 $E^\circ$ predicts only whether a reaction is thermodynamically feasible, not whether it will actually proceed at a useful rate. A positive $E^\circ_{\text{cell}}$ 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 $E^\circ_{\text{cell}}$ . 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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksUse the standard reduction potentials

\text{MnO}_4^-/\text{Mn}^{2+} = +1.51\ \text{V}

\text{Fe}^{3+}/\text{Fe}^{2+} = +0.77\ \text{V}

and

\text{I}_2/\text{I}^- = +0.54\ \text{V}

. (a) Predict whether acidified

\text{MnO}_4^-

will oxidise

\text{Fe}^{2+}

, justifying with

E^\circ_{\text{cell}}

. (b) Predict whether

\text{Fe}^{3+}

will oxidise

\text{I}^-

, justifying with

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

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

Positive, so the reaction is spontaneous: permanganate will oxidise

\text{Fe}^{2+}

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

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

Positive, so

\text{Fe}^{3+}

will oxidise

\text{I}^-

to iodine.

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

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

WACE 20234 marksExplain 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^\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 $\text{H}^+$ at $1\ \text{mol L}^{-1}$ , $\text{H}_2$ at $100\ \text{kPa}$ , $298\ \text{K}$ ) is assigned a potential of exactly $0\ \text{V}$ . Every other half-cell is measured against it, giving each a relative standard electrode potential.

$E^\circ$ 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 $E^\circ$ is never multiplied.

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