Predict spontaneity from cell potential and reduction tables

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Unit B2, Expectation 30-B2.7k

Drop a strip of zinc metal into a solution of copper(II) sulfate. Within seconds, the shiny grey zinc darkens, a reddish-brown solid appears on its surface, and the blue solution fades. Something is clearly happening — electrons are moving from zinc atoms to copper ions. This transfer of electrons is a redox reaction, and it is spontaneous: it happens on its own without any outside push of energy. But not every redox reaction is spontaneous. If you dropped copper metal into a zinc sulfate solution, nothing visible would happen. How can you predict — before running the experiment — which direction a redox reaction will go? The answer lies in a set of values called standard reduction potentials and a simple calculation of cell potential. This lesson builds that prediction tool from the ground up.

What you will learn

Use standard reduction potentials from the Alberta Chemistry 30 Data Booklet to identify the cathode (reduction) and anode (oxidation) half-reactions in an electrochemical cell.
Calculate the standard cell potential using the formula E°cell = E°cathode − E°anode.
Predict whether a redox reaction is spontaneous based on the sign of E°cell.
Apply the spontaneity prediction to both full cell reactions and individual redox reactions written as equations.

Prerequisite Review: Oxidation, Reduction, and Half-Reactions

Before using reduction potentials, recall two definitions. Oxidation is the loss of electrons by a species; reduction is the gain of electrons by a species. The memory aid OIL RIG — Oxidation Is Loss, Reduction Is Gain — is helpful. In every redox reaction, one species is oxidized while another is reduced at the same time.

A half-reaction isolates either the oxidation or the reduction step. For example, zinc losing two electrons is written as an oxidation half-reaction, and copper(II) ion gaining two electrons is written as a reduction half-reaction. When the two half-reactions are combined so that electrons cancel, the result is the overall balanced redox equation. This combining skill from Chemistry 20 is the foundation for everything in this lesson.

In an electrochemical cell, the two half-reactions are physically separated. The electrode where reduction occurs is the cathode. The electrode where oxidation occurs is the anode. Electrons flow through an external wire from the anode to the cathode.

Oxidation = loss of electrons; reduction = gain of electrons (OIL RIG).
A half-reaction shows only one of the two processes.
Cathode = reduction; anode = oxidation.
Electrons travel through the external circuit from anode to cathode.

Standard Reduction Potentials: Reading the Data Booklet Table

Every half-reaction has a measurable tendency to attract electrons. This tendency is expressed as a standard reduction potential, given the symbol E°. The unit is the volt (V). A larger (more positive) E° value means the species is a stronger oxidizing agent — it pulls electrons toward itself more forcefully.

The Alberta Chemistry 30 Data Booklet lists half-reactions written in the reduction direction, meaning electrons appear on the left side of each equation. For example, the half-reaction

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

has

E° = +0.34 \text{ V}

, and the half-reaction

\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)

has

E° = -0.76 \text{ V}

. Because copper(II) ion has the higher (more positive) E°, it is the stronger oxidizing agent and will be reduced in a cell that pairs these two half-reactions.

A species at the top-right of the reduction table (most positive E°) is the strongest oxidizing agent. A species at the bottom-left (most negative E°, written as the reverse, an oxidation) is the strongest reducing agent. This ranking directly tells you which direction reactions will proceed spontaneously.

Standard conditions mean a concentration of 1 mol/L for all aqueous species, a pressure of 101.3 kPa for any gases, and a temperature of 25°C. The word 'standard' signals these specific conditions; the superscript ° in E° is the notation for standard state.

E° in volts (V); more positive = stronger tendency to be reduced

Standard reduction potential E° measures the pull of a half-reaction for electrons, in volts.
The Data Booklet lists all half-reactions in the reduction direction.
More positive E° → stronger oxidizing agent → more likely to be reduced.
More negative E° → stronger reducing agent → more likely to be oxidized.
Standard conditions: 1 mol/L, 101.3 kPa, 25°C.

Calculating Standard Cell Potential

The standard cell potential,

E°_{\text{cell}}

, tells you the overall voltage a cell can produce. It is calculated from the two standard reduction potentials — one for each half-reaction — using a single formula from the Alberta Chemistry 30 Data Booklet.

The formula is:

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

. Here,

E°_{\text{cathode}}

is the standard reduction potential of the half-reaction occurring at the cathode (reduction), and

E°_{\text{anode}}

is the standard reduction potential of the same half-reaction written in the Data Booklet (still in the reduction direction, even though the reaction actually runs in reverse at the anode). You subtract rather than add because the anode reaction is the reverse of what is listed.

An important rule: do not multiply E° values by stoichiometric coefficients. Standard reduction potentials are intensive properties — they do not change when you double the amount of substance. If you need two moles of electrons to balance the equation and you double a half-reaction, the E° value stays the same.

The sign of

E°_{\text{cell}}

is the key to spontaneity. A positive

E°_{\text{cell}}

means the reaction is spontaneous under standard conditions. A negative

E°_{\text{cell}}

means the reaction is non-spontaneous under standard conditions — it would require an input of electrical energy to occur. An

E°_{\text{cell}}

of exactly zero means the system is at equilibrium.

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

Formula: $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ .
Both E° values are taken from the Data Booklet in the reduction direction.
Do NOT multiply E° values by stoichiometric coefficients.
Positive $E°_{\text{cell}}$ → spontaneous reaction.
Negative $E°_{\text{cell}}$ → non-spontaneous reaction.

Predicting Spontaneity from the Reduction Table Directly

You do not always need to calculate

E°_{\text{cell}}

numerically to predict spontaneity. The layout of the reduction table gives a shortcut based on relative positions.

In the standard reduction table, the half-reaction listed higher (more positive E°) will proceed in the forward direction (reduction) spontaneously when paired with a half-reaction listed lower (less positive or more negative E°), which will run in the reverse direction (oxidation). This is sometimes described as the 'diagonal rule': a spontaneous reaction pairs a species on the right side of a higher half-reaction (the oxidizing agent, to be reduced) with a species on the left side of a lower half-reaction (the reducing agent, to be oxidized).

For example,

\text{Cu}^{2+}(aq)

appears on the left of a higher half-reaction (

E° = +0.34 \text{ V}

), and

\text{Zn}(s)

appears on the right of a lower half-reaction (

E° = -0.76 \text{ V}

). Pairing them gives a spontaneous cell: copper(II) ions are reduced, and zinc is oxidized. This matches exactly what you see when zinc metal is placed in a copper(II) sulfate solution.

If you reverse the pairing — trying to reduce

\text{Zn}^{2+}(aq)

with

\text{Cu}(s)

— the half-reaction with the higher E° is now forced to run backwards. That gives a negative

E°_{\text{cell}}

, confirming non-spontaneity. No visible reaction occurs.

Spontaneous when E°_{cathode} > E°_{anode}, giving E°_{cell} > 0

Higher E° half-reaction runs forward (reduction) in a spontaneous cell.
Lower E° half-reaction runs in reverse (oxidation) in a spontaneous cell.
Spontaneous pairing: strong oxidizing agent (high E° $, left side) + strong reducing agent (low$ E°, right side).
Reversing a spontaneous cell pairing gives a non-spontaneous reaction.

Applying the Prediction: A Step-by-Step Strategy

Use the following strategy every time you are asked to predict whether a given redox reaction is spontaneous. First, identify the two species being oxidized and reduced in the question. Second, look up both of their half-reactions in the Data Booklet, written in the reduction direction, and record each E° value with its sign. Third, assign cathode (higher E°) and anode (lower E°). Fourth, apply the formula

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

and calculate the result. Fifth, interpret the sign: positive means spontaneous, negative means non-spontaneous.

When the question gives you a complete balanced redox equation and asks whether it is spontaneous as written, you must identify which species is being oxidized and which is being reduced in that specific equation. The oxidation states of atoms change; track those changes to assign roles correctly. Then follow the same five steps.

Always carry the sign of each E° value through your calculation. A common error is to treat a negative E° as positive because it 'looks like a large number.' The sign is chemically meaningful — it tells you the direction of the driving force.

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}; positive → spontaneous, negative → non-spontaneous

Step 1: Identify what is oxidized and what is reduced.
Step 2: Find both E° values from the Data Booklet.
Step 3: Assign cathode (higher E°) and anode (lower E°).
Step 4: Calculate $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ .
Step 5: Positive result → spontaneous; negative result → non-spontaneous.

Summary: How to Assign Cathode and Anode from the Reduction Table

Property	Cathode (Reduction)	Anode (Oxidation)
Process	Gains electrons	Loses electrons
E° value	Higher (more positive)	Lower (less positive or more negative)
Species role	Oxidizing agent	Reducing agent
Table position	Higher in the reduction table	Lower in the reduction table
Effect on E°cell	Added as positive term	Subtracted in formula

Worked example

Is the Reaction Between Iron(III) Ion and Tin Spontaneous?

A student mixes aqueous iron(III) ions with aqueous tin(II) ions. The proposed reaction is:

2\text{Fe}^{3+}(aq) + \text{Sn}^{2+}(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Sn}^{4+}(aq)

. Using standard reduction potentials from the Alberta Chemistry 30 Data Booklet, determine whether this reaction is spontaneous as written.

Step 1 — Identify oxidation and reduction
Iron changes from +3 to +2: it gains one electron each, so iron(III) ion is reduced. Tin changes from +2 to +4: it loses two electrons, so tin(II) ion is oxidized. Reduction occurs at the cathode; oxidation occurs at the anode.
$\text{Fe}^{3+} \rightarrow \text{Fe}^{2+} \text{ (reduction, cathode)}; \quad \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} \text{ (oxidation, anode)}$
Step 2 — Look up standard reduction potentials
Find both half-reactions in the Data Booklet, written in the reduction direction. Record the values exactly as listed, with their signs.
$\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E° = +0.77 \text{ V}$
Step 2 continued — second half-reaction
The tin half-reaction from the Data Booklet, written in the reduction direction, gives the anode value. Even though the reaction runs in reverse at the anode, always record the Data Booklet value as listed.
$\text{Sn}^{4+}(aq) + 2e^- \rightarrow \text{Sn}^{2+}(aq) \quad E° = +0.15 \text{ V}$
Step 3 — Assign cathode and anode
The half-reaction with the higher (more positive) standard reduction potential is the cathode. Iron(III)/iron(II) has E° = +0.77 V; tin(IV)/tin(II) has E° = +0.15 V. Therefore iron is the cathode and tin is the anode. This matches the direction given in the question.
$E°_{\text{cathode}} = +0.77 \text{ V} \quad (\text{Fe}^{3+}/\text{Fe}^{2+}); \quad E°_{\text{anode}} = +0.15 \text{ V} \quad (\text{Sn}^{4+}/\text{Sn}^{2+})$
Step 4 — Calculate E°cell
Apply the standard cell potential formula. Subtract the anode value from the cathode value. Do not multiply either E° value by any stoichiometric coefficient.
$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = +0.77 \text{ V} - (+0.15 \text{ V}) = +0.62 \text{ V}$
Step 5 — Interpret the sign
The calculated E°cell is positive (+0.62 V). A positive standard cell potential means the reaction proceeds spontaneously as written under standard conditions.
$E°_{\text{cell}} = +0.62 \text{ V} > 0 \Rightarrow \text{spontaneous}$

Answer: The reaction is spontaneous as written. The standard cell potential is

E°_{\text{cell}} = +0.62 \text{ V}

.

Check: Cross-check using the table position:

\text{Fe}^{3+}/\text{Fe}^{2+}

sits higher in the reduction table than

\text{Sn}^{4+}/\text{Sn}^{2+}

. So iron(III) is the stronger oxidizing agent and will be reduced spontaneously when paired with tin(II) as the reducing agent. The table-position prediction agrees with the calculated positive

E°_{\text{cell}}

.

Common mistakes and how to avoid them

Multiplying E° values by stoichiometric coefficients when balancing electrons. For example, doubling a half-reaction and writing 2 × (+0.77 V) = +1.54 V.

Correction: Standard reduction potentials are not multiplied by coefficients. E° is an intensive property — it does not depend on the amount of substance. Keep the Data Booklet value unchanged no matter how many times you multiply the half-reaction.

Subtracting the cathode from the anode instead of the anode from the cathode, reversing the sign of E°cell.

Correction: Always use E°cell = E°cathode − E°anode. Identify the cathode (higher E°) first, then subtract the anode value. Write the formula before plugging in numbers to avoid this reversal.

Ignoring the negative sign of a standard reduction potential. For example, treating E° = −0.76 V as +0.76 V in the subtraction.

Correction: Carry the sign of every E° value into the calculation. Subtracting a negative number adds its magnitude: E°cell = +0.34 V − (−0.76 V) = +1.10 V, not +0.34 V − 0.76 V = −0.42 V.

Concluding a reaction is spontaneous just because the reactants are familiar or because a reaction has been observed before, without calculating E°cell.

Correction: Always calculate E°cell from the Data Booklet values. Intuition about 'reactive' metals can be wrong for specific pairings. Let the sign of E°cell give the prediction.

Using the E° value for the oxidation direction instead of looking up the reduction direction from the Data Booklet, then forgetting to reverse the sign when applying the formula.

Correction: The Data Booklet lists all half-reactions in the reduction direction. Always read E° values directly from the table in that direction. The formula E°cell = E°cathode − E°anode automatically accounts for the reversal at the anode — you do not manually flip the sign of the anode E° value before substituting.

Lesson summary

Standard reduction potentials (E°, in volts) measure how strongly a half-reaction pulls electrons toward itself when written in the reduction direction.
The Data Booklet lists all half-reactions in the reduction direction; a more positive E° means a stronger oxidizing agent.
The cathode is always the half-reaction with the higher E°; the anode is the half-reaction with the lower E°.
Standard cell potential is calculated using E°cell = E°cathode − E°anode; E° values are never multiplied by stoichiometric coefficients.
A positive E°cell means the reaction is spontaneous under standard conditions; a negative E°cell means it is non-spontaneous.
The reduction table position provides a quick spontaneity check: a reaction is spontaneous when a species higher in the table (stronger oxidizing agent) oxidizes a species lower in the table (stronger reducing agent).

Check your understanding

Question 1

The standard reduction potentials for two half-reactions are:

\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)

,

E° = +0.80 \text{ V}

, and

\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s)

,

E° = -0.26 \text{ V}

. What is the standard cell potential for a cell that pairs these two half-reactions, with nickel as the anode?

$E°_{\text{cell}} = +0.54 \text{ V}$
$E°_{\text{cell}} = +1.06 \text{ V}$
$E°_{\text{cell}} = -1.06 \text{ V}$
$E°_{\text{cell}} = +0.80 \text{ V}$

Show answer and explanation

E°_{\text{cell}} = +1.06 \text{ V}

Nickel is the anode (lower E° = −0.26 V) and silver is the cathode (higher E° = +0.80 V). Applying the formula: E°cell = E°cathode − E°anode = +0.80 V − (−0.26 V) = +1.06 V. Subtracting a negative value adds its magnitude. Option A results from adding the magnitudes without attending to the negative sign. Options C and D involve incorrect formula arrangements.

Question 2

A student writes the following reaction:

\text{Cu}(s) + \text{Zn}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Zn}(s)

. The standard reduction potentials are

E°(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}

and

E°(\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V}

. Is this reaction spontaneous as written?

Yes, because copper is a metal and metals are reactive.
No, because E°cell = +0.34 V − (−0.76 V) = +1.10 V, which is positive.
No, because E°cell = −0.76 V − (+0.34 V) = −1.10 V, which is negative.
Yes, because E°cell = −0.76 V − (+0.34 V) = −1.10 V, which is negative.

Show answer and explanation

No, because E°cell = −0.76 V − (+0.34 V) = −1.10 V, which is negative.

As written, copper is oxidized (anode, E° = +0.34 V) and zinc(II) is reduced (cathode, E° = −0.76 V). Applying the formula: E°cell = E°cathode − E°anode = −0.76 V − (+0.34 V) = −1.10 V. A negative E°cell means the reaction is non-spontaneous as written. The reverse reaction (zinc oxidized, copper reduced) is the spontaneous direction.

Question 3

Which statement correctly describes how to use the standard reduction table to predict spontaneity without calculating E°cell numerically?

A spontaneous reaction pairs a species on the left of a lower half-reaction with a species on the right of a higher half-reaction.
A spontaneous reaction pairs a species on the left of a higher half-reaction (oxidizing agent) with a species on the right of a lower half-reaction (reducing agent).
Any reaction involving a metal with a negative E° is always spontaneous.
The half-reaction with the lower E° always acts as the cathode in a spontaneous cell.

Show answer and explanation

A spontaneous reaction pairs a species on the left of a higher half-reaction (oxidizing agent) with a species on the right of a lower half-reaction (reducing agent).

The species on the left side of a higher half-reaction is a stronger oxidizing agent and will be reduced (it acts as the cathode). The species on the right side of a lower half-reaction is a stronger reducing agent and will be oxidized (it acts as the anode). This pairing gives a positive E°cell and a spontaneous reaction. Option A reverses the positions. Options C and D are false generalizations.

Question 4

A half-reaction must be doubled to balance electrons in a cell. How does doubling the half-reaction affect its standard reduction potential, E°?

E° doubles because twice as many electrons are transferred.
E° is halved because the reaction is split into two steps.
E° stays the same because standard reduction potentials do not depend on the amount of substance.
E° becomes negative when the half-reaction is reversed for the anode.

Show answer and explanation

E° stays the same because standard reduction potentials do not depend on the amount of substance.

Standard reduction potentials are intensive properties — they describe the tendency per electron, not the total energy for a specific quantity. Multiplying a half-reaction by any coefficient does not change E°. Only when you reverse a half-reaction (switching it from cathode to anode) does the formula account for the change, through subtraction in E°cell = E°cathode − E°anode.

Key terms

Standard reduction potential (E°): The voltage measured for a half-reaction written in the reduction direction under standard conditions (1 mol/L, 101.3 kPa, 25°C). A more positive E° means a stronger tendency to gain electrons.
Standard cell potential (E°cell): The overall voltage of an electrochemical cell under standard conditions, calculated as E°cell = E°cathode − E°anode. Its sign predicts spontaneity.
Cathode: The electrode in an electrochemical cell where reduction (electron gain) occurs. It corresponds to the half-reaction with the higher standard reduction potential.
Anode: The electrode in an electrochemical cell where oxidation (electron loss) occurs. It corresponds to the half-reaction with the lower standard reduction potential.
Spontaneous reaction: A reaction that proceeds on its own without a continuous input of energy. In electrochemistry, a spontaneous redox reaction has a positive standard cell potential (E°cell > 0).
Oxidizing agent: The species that causes another species to be oxidized by accepting electrons itself. It is reduced in the reaction. In the reduction table, stronger oxidizing agents have more positive E° values.
Reducing agent: The species that causes another species to be reduced by donating electrons itself. It is oxidized in the reaction. In the reduction table, stronger reducing agents have more negative E° values.
Standard conditions: The reference conditions used when measuring and reporting standard reduction potentials: aqueous ion concentration of 1 mol/L, gas pressure of 101.3 kPa, and temperature of 25°C.

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30-B2.7k · Predict spontaneity from cell potential and reduction tables

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Unit B2, Expectation 30-B2.7k

What you will learn

Prerequisite Review: Oxidation, Reduction, and Half-Reactions

Standard Reduction Potentials: Reading the Data Booklet Table

Calculating Standard Cell Potential

Predicting Spontaneity from the Reduction Table Directly

Applying the Prediction: A Step-by-Step Strategy

Summary: How to Assign Cathode and Anode from the Reduction Table

Is the Reaction Between Iron(III) Ion and Tin Spontaneous?

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review