14.10-14.14 - Redox Feasibility And Disproportionation

Using Standard Cell EMF

The chemical question

Imagine placing a strip of iron into a solution containing copper(II) ions. A GCSE-style reactivity answer might say that iron is more reactive than copper, so iron should displace copper. At A level, the same prediction can be made from electron transfer data:

$\ce{Cu2+(aq) + 2e- <=> Cu(s)}$ — $E^{\circ} = +0.34\,\mathrm{V}$

$\ce{Fe2+(aq) + 2e- <=> Fe(s)}$ — $E^{\circ} = -0.44\,\mathrm{V}$

The half-equation with the more positive standard electrode potential is more likely to occur as a reduction. So $\ce{Cu2+}$ is predicted to gain electrons, while iron metal is oxidised:

reduction: $\ce{Cu2+(aq) + 2e- -> Cu(s)}$

oxidation: $\ce{Fe(s) -> Fe2+(aq) + 2e-}$

overall: $\ce{Fe(s) + Cu2+(aq) ->Fe2+(aq) + Cu(s)}$

To calculate the standard cell emf for the predicted direction, use the two values as standard reduction potentials:

A positive $E^{\circ}_{\mathrm{cell}}$ predicts that the reaction is thermodynamically feasible under standard conditions. It does not say the reaction is fast, complete or safe to carry out; it says the thermodynamic direction is favourable when the species are in their standard states.

Why the sign matters

For a redox reaction, the cell emf is linked to the maximum electrical work available from electron transfer. A positive $E^{\circ}_{\mathrm{cell}}$ means the chosen redox direction can release energy as electrical work under standard conditions. A negative $E^{\circ}_{\mathrm{cell}}$ means the reaction as written is not thermodynamically feasible under those conditions, although the reverse direction may be.

The sign also checks the chemistry. If the reverse of the iron/copper reaction were written:

then:

That negative value agrees with the chemical interpretation: copper metal is not predicted to reduce iron(II) ions to iron under standard conditions.

Common-error contrast

Do not multiply an $E^{\circ}$ value when a half-equation is multiplied to balance electrons. Electrode potential is an intensive value, like temperature: changing the amount of reaction changes the total energy transferred, but not the potential difference for that redox couple.

Cell EMF, Entropy and Equilibrium

The thermodynamic bridge

Electrode potentials are not a separate kind of feasibility rule. They connect to the same thermodynamic ideas used in entropy and Gibbs free energy.

For a reaction involving $n$ moles of electrons:

The important interpretation is:

The entropy sign in this table describes the reaction starting under standard conditions. At equilibrium the actual cell emf and actual total entropy change for further net reaction are zero. $E^{\circ}$ need not be zero at equilibrium: it remains the reference value linked to $K$. Also compare different reactions at the same $n$ and $T$, since the proportionality factor contains both.

This does not mean every positive reaction goes to completion. A positive value means products are favoured compared with reactants at equilibrium under the stated conditions.

Worked example: interpreting a positive value

A reaction has $E^{\circ}_{\mathrm{cell}} = +0.46\,\mathrm{V}$ under standard conditions.

Without calculating $K$, you can make three linked statements:

$E^{\circ}_{\mathrm{cell}}$ is positive.

So $\Delta S_{\mathrm{total}}$ for the reaction is positive.

So $\ln K$ is positive, which means $K > 1$.

Chemical interpretation: the products are thermodynamically favoured at equilibrium under standard conditions. The reaction may still need a suitable pathway or catalyst before a visible change is seen in the laboratory.

Worked example: interpreting a small negative value

A different reaction has $E^{\circ}_{\mathrm{cell}} = -0.03\,\mathrm{V}$.

The negative sign predicts that the reaction as written is not thermodynamically feasible under standard conditions:

$E^{\circ}_{\mathrm{cell}} < 0$

$\Delta S_{\mathrm{total}} < 0$

$\ln K < 0$

Because the value is close to zero, the prediction is especially sensitive to conditions. A change in concentration, pressure or pH could make the actual non-standard electrode potential different from the standard value.

Reading The Electrochemical Series

What the series ranks

Standard electrode potentials can be listed as an electrochemical series. In the Data Booklet, the half-equations are written as reductions. A small selection looks like this:

The more positive the value, the more strongly the left-hand species tends to gain electrons. So $\ce{Cl2}$ is a stronger oxidising agent than $\ce{Br2}$, and $\ce{Ag+}$ is a stronger oxidising agent than $\ce{Cu2+}$.

The more negative the value, the more readily the right-hand reduced form tends to lose electrons. So zinc metal is a stronger reducing agent than iron metal in this series.

Worked example: copper and silver ions

Predict whether copper metal reacts with silver ions under standard conditions.

Use the relevant reduction potentials:

$\ce{Ag+(aq) + e- <=>Ag(s)}$ — $E^{\circ} = +0.80\,\mathrm{V}$

$\ce{Cu2+(aq) + 2e- <=> Cu(s)}$ — $E^{\circ} = +0.34\,\mathrm{V}$

The more positive half-equation is reduction of $\ce{Ag+}$, so silver ions gain electrons. Copper metal must therefore be oxidised:

reduction: $\ce{2Ag+(aq) + 2e- -> 2Ag(s)}$

oxidation: $\ce{Cu(s) -> Cu2+(aq) + 2e-}$

overall: $\ce{Cu(s) + 2Ag+(aq) -> Cu2+(aq) + 2Ag(s)}$

Now calculate the standard cell emf: