Changing the Conditions Questions

1. The following cell is set up: Ni(s) | Ni²⁺(aq) || Cu²⁺(aq) | Cu(s)

a) State the conditions of the cell when run at standard conditions.

Pressure: 100 kPa
Temperature: 298 K
Ion Concentrations: 1.00 mol dm⁻³

b) Calculate the EMF of the cell.

EMF = E(Right) – E(Left)
= +0.34 – (-0.25) = +0.59 V

c) State the effect of increasing the size of the nickel electrode on the EMF.

No effect. The surface area of the solid electrode does not affect the standard electrode potential.

d) State and explain the effect of increasing the concentration of Cu²⁺.

The equilibrium is: Cu²⁺ + 2e⁻ ⇌ Cu.
Increasing [Cu²⁺] shifts the position of equilibrium to the right (reduction is favoured). This makes the potential of the copper half-cell more positive (higher). As Cu is the cathode (right-hand side), increasing its potential increases the difference between the two electrodes, resulting in a larger EMF.

2. The following cell is set up: Pt(s) | H₂(g) | H⁺(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s)

a) State the conditions of the S.H.E.

Pressure: 100 kPa
Temperature: 298 K
[H⁺]: 1.00 mol dm⁻³

b) Calculate the EMF of the cell.

EMF = +0.77 – 0.00 = +0.77 V

c) State and explain the effect of increasing the concentration of H⁺.

The S.H.E equilibrium is: 2H⁺ + 2e⁻ ⇌ H₂.

Increasing [H⁺] shifts equilibrium to the right, making the potential more positive (higher than 0.00V). Since the S.H.E is the anode (left side), increasing its potential reduces the difference between the cathode and anode (EMF = E_cathode – E_anode).
Result: Smaller EMF.

d) State the effect of increasing the temperature of the hydrogen half-cell. (Note: Reduction of H⁺ is exothermic).

Increasing temperature favours the endothermic direction (backwards/oxidation). This shifts equilibrium to the left, making the potential more negative (lower). Since the S.H.E is the anode, decreasing its potential increases the gap between it and the cathode.
Result: Larger EMF.

e) State and explain the effect of decreasing the concentration of Fe³⁺.

Equilibrium: Fe³⁺ + e⁻ ⇌ Fe²⁺.

Decreasing [Fe³⁺] shifts equilibrium to the left. This makes the potential less positive (lower). Since this is the cathode, lowering its potential reduces the difference between electrodes.
Result: Smaller EMF.

3. The following concentration cell is set up: Cu(s) | Cu²⁺(aq, 0.5M) || Cu²⁺(aq, 1.0M) | Cu(s)

a) Write the half equations for both cells and state the direction of electron flow.

Left (Anode): Cu ⇌ Cu²⁺ + 2e⁻ (Oxidation)

Right (Cathode): Cu²⁺ + 2e⁻ ⇌ Cu (Reduction)

Electrons flow from Left to Right (Low conc to High conc).

b) Predict the concentration in the left half-cell when the cell stops generating electricity.

0.75 mol dm⁻³ (The concentrations will equalise).

c) A new cell is made with equal concentrations. Right cell at 50°C, Left cell at 25°C. Predict electron flow direction.

Given reduction is endothermic, higher temperature (Right) favours reduction. This makes the Right potential higher (more positive) than the Left.

Therefore, the Right cell acts as the Cathode (Reduction) and the Left cell as the Anode (Oxidation).

Electrons flow from Left to Right.