Galvanic Cell: Concentration Effects and the Cost of a Discharge
A student builds the cell Zn(s) | Zn²⁺(0.10 M) || Cu²⁺(0.010 M) | Cu(s) at 298 K, using 100 mL of each solution and a saturated KNO₃ salt bridge. Standard electrode potentials are E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V. The cell is discharged for 10 minutes at an average current of 0.20 A, after which the student re-measures. In the write-up the EMF is recorded as exactly +1.10 V with the concentrations ignored, and the cell diagram is written with zinc on the right and copper on the left. The demonstrator marks both as errors and also asks whether adding concentrated ammonia to the copper compartment, which forms [Cu(NH₃)₄]²⁺ with log β = 13, would raise or lower the measured EMF. Assume the only effect of the ammonia is to lower the free Cu²⁺ concentration and that the temperature stays at 298 K.
1. Calculate the actual EMF of the cell as built, identify the anode and cathode, and write the cell diagram in the accepted convention.
Read the worked answer
Standard cell potential: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (+0.34) - (-0.76) = 1.10\ \text{V}.$$
Zinc has the more negative potential, so zinc is oxidised: Zn is the anode (negative electrode), Cu is the cathode.
Nernst equation for $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, with $n = 2$:
$$E = E^\circ - \frac{0.0592}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 1.10 - 0.0296\log\frac{0.10}{0.010}.$$
$$\log\frac{0.10}{0.010} = \log 10 = 1$$
$$E = 1.10 - 0.0296 = 1.07\ \text{V}.$$
So the recorded 1.10 V is wrong by 30 mV because the solutions are not at unit activity; the ratio being 10:1 in the 'wrong' direction (more Zn²⁺, less Cu²⁺) makes the cell slightly weaker than standard. The accepted notation puts the anode on the left, reduced species first:
$$\text{Zn(s)} \mid \text{Zn}^{2+}(0.10\ \text{M}) \parallel \text{Cu}^{2+}(0.010\ \text{M}) \mid \text{Cu(s)}.$$
The student's version, Cu|Cu²⁺||Zn²⁺|Zn, describes the reverse (non-spontaneous) cell and would correspond to $-1.07$ V if quoted with the same sign convention.
2. Does adding concentrated ammonia to the copper compartment raise or lower the EMF? Justify with the Nernst equation, and estimate the size of the effect.
Read the worked answer
It lowers the EMF.
Ammonia sequesters $\text{Cu}^{2+}$ as $[\text{Cu(NH}_3)_4]^{2+}$ with $\log\beta = 13$, so the free $[\text{Cu}^{2+}]$ falls sharply while $[\text{Zn}^{2+}]$ is unchanged. In
$$E = 1.10 - 0.0296\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]},$$
a smaller $[\text{Cu}^{2+}]$ makes the ratio larger, so the logarithm grows and $E$ decreases. Physically, the cathode's tendency to be reduced is weakened when its ions are stabilised in a complex, so the driving force falls.
Order of magnitude: if complexation lowers free $\text{Cu}^{2+}$ by a factor of about $10^{10}$ (a realistic figure with a large excess of ammonia), the log term increases by 10 and
$$\Delta E = -0.0296\times10 = -0.30\ \text{V},$$
so the EMF would drop from about 1.07 V to roughly 0.77 V. Note the sign: decreasing the cathode ion concentration always reduces the cell potential, whereas decreasing the anode ion concentration would increase it.
3. After 10 minutes at an average of 0.20 A, by how much has the EMF fallen? Use 1 F = 96500 C mol⁻¹.
Read the worked answer
Charge passed: $Q = It = 0.20\times600 = 120$ C.
Moles of electrons: $n_e = 120/96500 = 1.24\times10^{-3}$ mol.
For $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$, 2 mol electrons per mol Zn, so
$$\Delta n(\text{Zn}^{2+}) = +\frac{1.24\times10^{-3}}{2} = +6.2\times10^{-4}\ \text{mol}$$ $$\Delta n(\text{Cu}^{2+}) = -6.2\times10^{-4}\ \text{mol}.$$
In 0.100 L:
- $[\text{Zn}^{2+}] = 0.10 + 0.0062 = 0.106$ M
- $[\text{Cu}^{2+}] = 0.010 - 0.0062 = 0.0038$ M
$$\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.106}{0.0038} = 28$$
$$E = 1.10 - 0.0296\log 28 = 1.10 - 0.0296(1.45) = 1.10 - 0.043 = 1.06\ \text{V}.$$
Combined with the initial correction, the EMF has fallen from about 1.07 V to about 1.06 V in ten minutes - only ~13 mV. Judgement: the cell is far from exhausted even though a visible amount of copper has been consumed, because the EMF depends logarithmically on concentrations; the cell will 'die' abruptly by polarisation and depletion long before the Nernst term alone predicts a collapse. The mass change at the electrodes (about 40 mg of Zn dissolved and 39 mg of Cu deposited) is a better indicator of how much charge has been delivered.