Two-Metal Galvanic Cell — Nernst Equation: 16-Chem-B12, Undated paper, Q1

Question 1: Two-Metal Galvanic Cell — Nernst Equation (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

The question is read as a copper–lead cell: a Cu cathode at $[\text{Cu}^{2+}]=0.30\ \text{M}$, a Pb anode (oxidized), cell potential $0.507\ \text{V}$, solve for $[\text{Pb}^{2+}]$. That reading reconciles “0.3 molar Cu²⁺”, “lead electrode is oxidized” and “0.507 V”, and is solved below.

Given. A two-electrode galvanic cell, Pb (anode, oxidised) and Cu (cathode, reduced), in solutions of their divalent ions at $T=25^{\circ}\text{C}$. The copper half-cell is at $[\text{Cu}^{2+}]=0.30\ \text{M}$ and the measured cell potential is $E_{cell}=0.507\ \text{V}$.

Quantity	Value
Standard potential, Cu²⁺ + 2e⁻ → Cu	$E^{\circ}_{Cu}=+0.337\ \text{V}$
Standard potential, Pb²⁺ + 2e⁻ → Pb	$E^{\circ}_{Pb}=-0.126\ \text{V}$
Cu²⁺ concentration	$0.30\ \text{M}$
Measured cell potential	$0.507\ \text{V}$
Electrons transferred, $n$	$2$

Find. The lead-ion concentration $[\text{Pb}^{2+}]$ consistent with the measured cell potential.

Figure 1.1 — The galvanic cell. Lead is the anode (Pb → Pb²⁺ + 2e⁻); electrons flow through the external wire to the copper cathode (Cu²⁺ + 2e⁻ → Cu); the salt bridge carries ionic current. The voltmeter reads the working cell potential 0.507 V.

Approach

Write the spontaneous cell reaction, get the standard cell EMF from the EMF series, then apply the Nernst equation and solve the resulting reaction quotient for the unknown lead-ion activity.

Identify the cell reaction and standard EMF. Lead is stated to oxidise (anode) and copper to reduce (cathode), so $$\text{Pb} + \text{Cu}^{2+} \longrightarrow \text{Pb}^{2+} + \text{Cu}$$ The standard cell potential is the cathode minus the anode standard potential: $$E^{\circ}_{cell} = E^{\circ}_{Cu} - E^{\circ}_{Pb} = 0.337 - (-0.126) = 0.463\ \text{V}$$ ==**$E^{\circ}_{cell} = 0.463$ V**== (positive ⇒ the reaction as written is spontaneous, consistent with Pb being the anode).
Write the Nernst equation. For the two-electron reaction, with the reaction quotient $Q=[\text{Pb}^{2+}]/[\text{Cu}^{2+}]$ (pure solids have unit activity) and the 25 °C base-10 form $2.303RT/F = 0.0592\ \text{V}$: $$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n}\,\log_{10}\!\frac{[\text{Pb}^{2+}]}{[\text{Cu}^{2+}]}$$
Insert the measured potential and solve for the log term. With $n=2$, $E_{cell}=0.507$ V: $$0.507 = 0.463 - \frac{0.0592}{2}\,\log_{10}\!\frac{[\text{Pb}^{2+}]}{0.30}$$ $$\log_{10}\!\frac{[\text{Pb}^{2+}]}{0.30} = \frac{0.463-0.507}{0.0296} = -1.486$$
Recover the lead-ion concentration. Taking antilogs, $$\frac{[\text{Pb}^{2+}]}{0.30} = 10^{-1.486} = 0.0326 \quad\Rightarrow\quad [\text{Pb}^{2+}] = 0.30 \times 0.0326$$ ==**$[\text{Pb}^{2+}] \approx 9.8 \times 10^{-3}\ \text{M}$**==

The measured potential (0.507 V) exceeds the standard value (0.463 V), so the cell is driven harder than standard; by Le Chatelier this requires the product ion Pb²⁺ to be dilute relative to the reactant Cu²⁺ — exactly the small concentration obtained.

Quantity	Value
Standard cell EMF $E^{\circ}_{cell}$	$0.463\ \text{V}$
Reaction quotient $[\text{Pb}^{2+}]/[\text{Cu}^{2+}]$	$0.0326$
Lead-ion concentration $[\text{Pb}^{2+}]$	$9.8\times10^{-3}\ \text{M}$

Concept & Theory

Topic. Galvanic-cell thermodynamics and the Nernst equation.

Key relations. $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$; Nernst $E=E^{\circ}-\dfrac{0.0592}{n}\log_{10}Q$ at 25 °C, with $Q$ built from ion activities (pure metals = 1).

Why this works. Each electrode potential shifts with ion activity according to the Nernst relation; the cell reads the difference of the two half-cell potentials. Because the log coefficient is only $0.0592/n$ volts per decade, a modest 44 mV departure from the standard EMF corresponds to nearly a decade-and-a-half change in the concentration ratio — which is why the lead ion must be so dilute.

Common pitfall. Reversing the quotient (putting Cu²⁺ over Pb²⁺), or forgetting that oxidation makes Pb the anode so its ion is the product in $Q$. A second trap is using $0.0592$ instead of $0.0592/n$ — the divalent ions make $n=2$.

Source. Fontana, Corrosion Engineering 3rd ed., Ch. 9 (electrode potentials, EMF series, Nernst equation); Jones, Principles and Prevention of Corrosion Ch. 2.

Question 1 of 8: Two-Metal Galvanic Cell — Nernst Equation

Question 1: Two-Metal Galvanic Cell — Nernst Equation (10 marks)

Approach