Topic 16 · page 11 of 19

Limits of predictions from standard potentials

A positive emf says a reaction can happen, not that it will happen quickly, and the value only applies under standard conditions. Learn the two limits and the evidence for each.

11 min read16.12

Read first: Cell emf, entropy and the equilibrium constant

Kinetic stability 16.12

$E^\ominus_\text{cell}$ tells you whether a reaction is feasible. It says nothing about how fast the reaction goes. If the activation energy is high, the reaction can be far too slow to see, even though it is feasible.

The standard example is hydrogen with oxygen. The two couples are O₂/H₂O (+1.23 V) and H⁺/H₂ (0.00 V):

E^\ominus_\text{cell} = +1.23 - 0.00 = +1.23\ \text{V}

The reaction is strongly feasible. Yet a mixture of hydrogen and oxygen sits unchanged at room temperature until it is sparked or a catalyst is added. The mixture is kinetically stable but thermodynamically unstable.

Departure from standard conditions 16.12

Standard electrode potentials apply at 1.00 mol dm⁻³, 100 kPa and 298 K. Real reactions are rarely run like that, and, as page 8 showed, a change in conditions changes E. If the value of $E^\ominus_\text{cell}$ is close to zero, a change of conditions can reverse the prediction.

Worked exampleChlorine from manganese(IV) oxide

$E^\ominus$ (MnO₂/Mn²⁺) = +1.23 V and $E^\ominus$ (Cl₂/Cl⁻) = +1.36 V.

MnO₂ is reduced and Cl⁻ is oxidised.
$E^\ominus_\text{cell}$ = +1.23 − (+1.36) = −0.13 V. Not feasible under standard conditions.
Yet chlorine is made in the lab by warming MnO₂ with concentrated hydrochloric acid.

\begin{aligned} &\ce{MnO2(s) + 4HCl(aq)} \\ &\ce{-> MnCl2(aq) + Cl2(g) + 2H2O(l)} \end{aligned}

The explanation uses the two equilibria.

The two electrode equilibria are:

\begin{aligned} &\ce{MnO2(s) + 4H+(aq) + 2e-} \\ &\ce{<=> Mn^2+(aq) + 2H2O(l)} \end{aligned}

\ce{Cl2(g) + 2e- <=> 2Cl-(aq)}

A high [H⁺] shifts the first equilibrium to the right, so $E$ (MnO₂/Mn²⁺) becomes more positive.
A high [Cl⁻] shifts the second equilibrium to the left, so $E$ (Cl₂/Cl⁻) becomes less positive.
The gap flips and $E_\text{cell}$ becomes positive. Heating, and chlorine escaping as a gas, help too.

Check yourself

State the two limitations of predictions made with standard electrode potentials.
Hydrogen and oxygen can be stored together as a mixture even though $E^\ominus_\text{cell}$ = +1.23 V. Explain why.
Calculate $E^\ominus_\text{cell}$ for MnO₂ with Cl⁻ in acid and say what it predicts. Why is chlorine nevertheless made from MnO₂ and hydrochloric acid?
A student predicts that a reaction is feasible but sees no change. Give a likely explanation.
A reaction has $E^\ominus_\text{cell}$ = −0.05 V but occurs when one reactant is used in much higher concentration than 1.00 mol dm⁻³. Explain why this is possible.

Answers cover these until you have tried the questions

(i) Kinetic stability: a feasible reaction may be too slow to happen because the activation energy is high. (ii) Departure from standard conditions: $E^\ominus$ values apply only at 298 K, 100 kPa and 1.00 mol dm⁻³, and real conditions change E.
The reaction has a high activation energy, so it is too slow at room temperature. The mixture is kinetically stable, even though the reaction is feasible.
$E^\ominus_\text{cell}$ = +1.23 − (+1.36) = −0.13 V, so it is predicted not to be feasible. With concentrated HCl, the high [H⁺] makes E(MnO₂/Mn²⁺) more positive and the high [Cl⁻] makes E(Cl₂/Cl⁻) less positive, so $E_\text{cell}$ becomes positive.
The reaction has a high activation energy, so the rate is too slow: the system is kinetically stable.
The value is only slightly negative. A much higher concentration of a reactant shifts its electrode equilibrium and makes the cell emf more positive, so $E_\text{cell}$ can become positive. Conditions are not standard.