Electrochemistry: Question 5

Worked solution

Step 1: Identify which way the equilibrium shifts on dilution

The electrode equilibrium is:

$\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)$

Decreasing $[\text{Cu}^{2+}(aq)]$ (by adding water) removes a species from the left-hand side. By Le Chatelier’s principle, the system responds by shifting to partially replenish the depleted $\text{Cu}^{2+}(aq)$. That is, the equilibrium shifts to the left, favouring the reverse process $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$.

Step 2: Relate the shift to the electrode potential

Shifting left means the electrode surface effectively accumulates relatively more free electrons (the copper is, comparatively, releasing electrons rather than the solution supplying $\text{Cu}^{2+}$ to accept them). This corresponds to a weaker tendency for the forward, electron-gaining (reduction) reaction to occur relative to standard conditions, i.e. the half-cell’s tendency to be reduced is diminished. A weaker tendency to be reduced means a less positive (lower) electrode potential.

This qualitative conclusion matches the quantitative Nernst equation for this couple, $E = E^{\ominus} + \dfrac{RT}{2F}\ln[\text{Cu}^{2+}]$: as $[\text{Cu}^{2+}]$ decreases, $\ln[\text{Cu}^{2+}]$ becomes more negative, so $E$ falls below $E^{\ominus}$.

Step 3: Confirm this is the only fully correct option

• A: gets both the direction of the shift (should be left, not right) and the resulting effect (should be less positive, not more positive) wrong.
• B: correctly shifts left and correctly concludes the electrode potential becomes less positive, this matches the reasoning above.
• C: incorrect. Only the standard electrode potential (measured at exactly $1.00\ \text{mol dm}^{-3}$) is a fixed reference value; the actual electrode potential of a half-cell does change when its concentration is no longer standard.
• D: reaches the correct conclusion (less positive) but by self-contradictory reasoning. A shift to the right would mean more $\text{Cu}^{2+}$ is being reduced, which would make the potential more positive, not less.

Final answer

• The electrode potential becomes less positive, because diluting $\text{Cu}^{2+}(aq)$ shifts the equilibrium to the left, reducing the tendency of the half-cell to gain electrons, option B.