Topic 16 · page 10 of 19

Cell emf, entropy and the equilibrium constant

A positive cell emf means a positive total entropy change, which is why the reaction is feasible. The emf is directly proportional to the total entropy change and to ln K, so it also tells you how far the reaction goes.

11 min read16.11

Read first: Predicting feasibility

Why a positive emf means feasible 16.11

Page 9 gave a rule: positive $E^\ominus_\text{cell}$ means feasible. The reason comes from entropy. A reaction is feasible when the total entropy change is positive. A cell that pushes electrons round a circuit is doing exactly that: the more strongly it pushes, the more the total entropy rises.

The spec states the link as two proportionalities.

K is the equilibrium constant. A large K means the equilibrium lies far to the right, towards products. So $E^\ominus_\text{cell}$ , $\Delta S_\text{total}$ and ln K all rise and fall together.

A graph of ln K, and of total entropy change, against E cell. It is a straight line through the origin. Points with positive E cell, such as a zinc-copper cell at plus 1.10 volts, are feasible with K greater than one. Points with negative E cell, such as iron(III) with bromide at minus 0.32 volts, have K less than one. — Directly proportional means a straight line through the origin. Positive E cell: feasible, K > 1. Negative E cell: not feasible, K < 1.

Reading the emf as an extent 16.11

$E^\ominus_\text{cell}$	$\Delta S_\text{total}$	K	What happens
large and positive	large and positive	much greater than 1	goes essentially to completion
small and positive	small and positive	just above 1	an equilibrium mixture, with much reactant left
zero	zero	1	no net change
negative	negative	less than 1	not feasible; the reverse reaction is

Two cells from this topic show the two ends of the table.

Zinc with copper(II): $E^\ominus_\text{cell}$ = +1.10 V. Large and positive, so K is huge and zinc removes almost all the copper(II) ions from solution.
Silver ions with iron(II) ions: $E^\ominus_\text{cell}$ = +0.03 V. The reaction is feasible, but the value is so small that K is only just above 1. A lot of reactant is left at equilibrium.

Worked exampleComparing two reactions

Reaction A has $E^\ominus_\text{cell}$ = +0.74 V and reaction B has +0.23 V. Which goes further towards completion, and what can you say about each K?

$E^\ominus_\text{cell}$ is directly proportional to ln K, and both are positive.
The larger value, +0.74 V, gives the larger ln K.
Reaction A goes further. Both have K > 1 (feasible), but K for A is much larger than for B.

Check yourself

State how $E^\ominus_\text{cell}$ is related to the total entropy change and to ln K.
A reaction has $E^\ominus_\text{cell}$ = +1.50 V. Describe the likely position of equilibrium.
A reaction has $E^\ominus_\text{cell}$ = +0.05 V. What does this tell you about K and about the mixture at equilibrium?
A reaction has $E^\ominus_\text{cell}$ = −0.40 V. What can you say about $\Delta S_\text{total}$ and K?
Explain why a positive $E^\ominus_\text{cell}$ means a reaction is feasible, using entropy.

Answers cover these until you have tried the questions

$E^\ominus_\text{cell}$ is directly proportional to $\Delta S_\text{total}$ and to ln K.
$E^\ominus_\text{cell}$ is large and positive, so K is much greater than 1 and the reaction goes essentially to completion: the equilibrium lies far to the right.
The value is small and positive, so ln K is small and positive and K is only just above 1. The reaction is feasible, but the equilibrium mixture contains a lot of unreacted reactants.
Negative $E^\ominus_\text{cell}$ means $\Delta S_\text{total}$ is negative and K is less than 1. The reaction is not feasible; the reverse reaction is.
$E^\ominus_\text{cell}$ is proportional to $\Delta S_\text{total}$ . A positive $E^\ominus_\text{cell}$ therefore means a positive total entropy change, and reactions are feasible when the total entropy change is positive.