Chapter 6: Equilibria

Solve · page 95: 2. Questions for Short Answers ii. How the following reversible reaction will be affected if its temperature is increased? Electricity

Effect of raising the temperature

1. Identify the heat term of the reversible reaction

The redox reaction that produces electricity in a galvanic cell is exothermic (heat is released):

$$\Delta H^\circ < 0$$

1. Apply Le Chatelier’s principle

When the temperature of a system at equilibrium is increased, the system tries to absorb the added heat.

• For an exothermic reaction the heat is a product; therefore the equilibrium shifts to the left (toward the reactants).

1. Consequences for the cell

Equilibrium constant – from the van’t Hoff equation

$$\ln K = -\frac{\Delta H^\circ}{R}\frac{1}{T}+\frac{\Delta S^\circ}{R}$$

because $\Delta H^\circ$ is negative, raising \(T\) makes \(\ln K\) smaller, so \(K\) decreases.

Cell potential – from the Nernst equation (at standard conditions)

$$E = \frac{RT}{nF}\ln K$$

a smaller \(K\) gives a smaller EMF (\(E\)).

1. Result

• The equilibrium moves toward the reactants.
• Less electricity (lower cell voltage) is produced.

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Answer: Increasing the temperature shifts the reversible (exothermic) reaction toward the reactants, thereby decreasing the amount of electricity generated.

Check: Use the van’t Hoff equation to see that \(K\) falls when \(T\) rises for a negative \(\Delta H^\circ\); then insert the new \(K\) into the Nernst equation and verify that the cell potential \(E\) becomes smaller.