A Daniell cell that gives the wrong voltage after dilution
A student assembles a Daniell cell: Zn(s)|Zn2+(1.0 M)||Cu2+(1.0 M)|Cu(s), using a porous pot instead of a salt bridge. A voltmeter reads 1.08 V at 298 K. The student then dilutes the copper half-cell to 0.010 M Cu2+ with distilled water, keeps the zinc half-cell at 1.0 M, and records 1.14 V. Standard potentials are E°(Zn2+/Zn) = −0.76 V and E°(Cu2+/Cu) = +0.34 V. The teacher says the second reading cannot be correct if the cell is wired and labelled properly. After 30 minutes the copper electrode has gained a reddish deposit and the blue colour of the Cu2+ solution has faded. The group must write the cell reaction, calculate the expected EMF after dilution, and decide whether the fault is in the wiring, the label, or the measurement.
1. Write the cell reaction, calculate E°cell, ΔG° and the equilibrium constant K at 298 K.
Read the worked answer
At the anode, zinc is oxidised:
[ \mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^-} ]
At the cathode, copper(II) is reduced:
[ \mathrm{Cu^{2+}(aq)+2e^-\rightarrow Cu(s)} ]
Overall:
[ \mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)} ]
[ E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}} =0.34-(-0.76)=1.10\ \mathrm{V} ]
[ \Delta G^\circ=-nFE^\circ=-2(96485)(1.10) \approx-2.12\times10^5\ \mathrm{J\ mol^{-1}} ]
The negative ΔG° confirms spontaneity. Using (E^\circ=(0.0591/n)\log K):
[ \log K=\frac{2(1.10)}{0.0591}\approx37.2 ]
[ K\approx10^{37} ]
The measured 1.08 V at 1.0 M is close to the standard value; the small difference is consistent with liquid-junction effects from the porous pot and with ordinary measurement error.
2. Use the Nernst equation to calculate E when [Cu2+] = 0.010 M and [Zn2+] = 1.0 M. Compare this with the reported 1.14 V and diagnose the likely error.
Read the worked answer
For the overall reaction,
[ Q=\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} =\frac{1.0}{0.010}=100 ]
[ E=E^\circ-\frac{0.0591}{2}\log Q =1.10-\frac{0.0591}{2}(2) ]
[ E=1.10-0.0591=1.04\ \mathrm{V} ]
Diluting Cu2+ lowers the cathode potential, so the cell EMF must fall from about 1.10 V to about 1.04 V, not rise to 1.14 V. The reported 1.14 V is higher than the standard value, which would require (Q<1), i.e. [Zn2+] < [Cu2+]. That is the opposite of what the student says was done. The likely errors are: the zinc half-cell was diluted instead of the copper half-cell, the half-cells were mislabelled, or the voltmeter leads were reversed (though reversal would normally give a negative reading). The porous pot can cause a small liquid-junction error, but not a 0.10 V error in this direction.
3. The reddish deposit and fading blue colour show that copper is being deposited. If a current of 0.200 A flows for 30.0 min, calculate the mass of copper deposited and state whether this is consistent with the observed colour change.
Read the worked answer
Quantity of charge:
[ Q=It=(0.200\ \mathrm{A})(30.0\times60\ \mathrm{s})=360\ \mathrm{C} ]
Moles of electrons:
[ n_e=\frac{360}{96485}=3.73\times10^{-3}\ \mathrm{mol} ]
Since (\mathrm{Cu^{2+}+2e^-\rightarrow Cu}), moles of Cu = (n_e/2=1.87\times10^{-3}) mol.
[ m_{\mathrm{Cu}}=(1.87\times10^{-3})(63.5)=0.119\ \mathrm{g} ]
This is a small but real deposit. It is consistent with copper(II) being reduced at the cathode and with the blue colour fading, because Cu2+(aq) is the species responsible for the blue colour. The observation also confirms that the copper electrode is the cathode under the normal Daniell-cell arrangement; if copper were dissolving instead, the blue colour would intensify and the electrode would lose mass.