Identifying an unknown metal from a cell potential
A student constructs a galvanic cell at 298 K from an unknown metal M dipping into 0.0100 mol/L M(NO3)2 solution and a silver electrode dipping into 0.100 mol/L AgNO3 solution. A potassium nitrate salt bridge connects the half-cells. The measured cell emf is 0.460 V, and the silver electrode is the positive terminal. The student's data table lists standard reduction potentials: Ag+/Ag +0.80 V; Cu2+/Cu +0.34 V; Pb2+/Pb -0.13 V; Ni2+/Ni -0.26 V; Zn2+/Zn -0.76 V. The student concludes that M is copper and writes the cell diagram. A second student points out that the silver nitrate solution may have been prepared from a stock bottle that had stood open, so its concentration might be lower than labelled.
1. Calculate the standard reduction potential of M2+/M from the measured emf and identify M. Show the Nernst calculation explicitly.
Read the worked answer
For the cell, Ag is the cathode because it is the positive terminal, and M is the anode. The cell reaction is
M(s) + 2Ag+ -> M2+ + 2Ag(s).
The reaction quotient is Q = [M2+]/[Ag+]^2 = 0.0100/(0.100)^2 = 1.00, so log Q = 0.
The Nernst equation is E = E° - (0.0592/2)log Q. With log Q = 0, E = E° = 0.460 V.
But E°cell = E°cathode - E°anode = 0.80 - E°(M2+/M). Therefore E°(M2+/M) = 0.80 - 0.460 = 0.34 V. The table value closest to this is Cu2+/Cu at +0.34 V, so M is copper. The identification is internally consistent with the measured emf and the stated concentrations.
2. Write the cell diagram and overall cell reaction. Calculate the emf expected if the Ag+ concentration were actually 0.0500 mol/L, and assess whether the identification would change.
Read the worked answer
Cell diagram: M(s) | M2+(0.0100 mol/L) || Ag+(0.100 mol/L) | Ag(s). Overall reaction: M(s) + 2Ag+ -> M2+ + 2Ag(s).
If [Ag+] = 0.0500 mol/L, then Q = 0.0100/(0.0500)^2 = 4.00.
E = 0.460 - (0.0592/2)log(4.00)
E = 0.460 - (0.0296)(0.602) = 0.460 - 0.0178 = 0.442 V.
The measured 0.460 V would then be 0.018 V higher than expected for copper. That is a small discrepancy, about 4%, and it would not bring any other metal in the table close: Pb would give about 0.93 V, Ni about 1.06 V and Zn about 1.56 V under the same conditions. The identification as copper remains reasonable within the precision of the experiment, but the concentration error should be acknowledged.
3. Explain how contamination of the silver half-cell by chloride, forming AgCl, would affect the measured emf and the conclusion.
Read the worked answer
Chloride would precipitate silver ions as AgCl, lowering the free [Ag+] in the silver half-cell. According to the Nernst equation, a lower [Ag+] decreases the cathode potential and therefore decreases the measured cell emf. For example, if free [Ag+] fell from 0.100 to 0.0500 mol/L, the emf would fall by about 0.018 V. A larger decrease would make the measured emf lower than the value expected for copper.
If the student ignored the chloride contamination, they might calculate an apparent E° for M that is too low, because they would subtract the depressed cathode potential from the measured emf. This could shift the identification toward a metal with a lower standard potential. The correct judgement is that the silver half-cell must be free of halide ions, and the Ag+ concentration must be known, before the cell emf can be used to identify M reliably.