Note that, defined in this way, percent differences can be positiveor negative.
ID: 75608 • Letter: N
Question
Note that, defined in this way, percent differences can be positiveor negative.
Cell Measured Potential Calculated Potential % difference Ag-Zn V V Pb-Zn V V Cu-Zn V V
Explanation / Answer
The error you made was multiplying the potential of the Ag+ + 1e-> Ag by 2 You don't have to do that. Potentials are not multiplied when youchange the stoichiometric coefficients. For a half cell, E = E0- RT/nF *ln(activity) If you multiply the equation by 2, 2Ag+ + 2e- > 2Ag E = E0- RT/(nF)*ln{1/[Ag+]^2} n = 2 electrons. The power in the logarithm comes out and cancelsthe n = 2. E = E0 - RT/(2F)*2*ln{1/[Ag+]} = E0 - RT/F *ln{1/[Ag+]} If you kept it as Ag+ + 1e- > Ag E = E0 - RT/(nF)*ln{1/[Ag+]}, but n = 1, and you get the same expression. Thus you should have taken V = 0.8 - (-0.76) = 1.56, and not 1.6 - (-0.76) Also the % difference Ag-Zn : (1.56-1.142)/1.56 * 100% = 26.8% Pb-Zn: (0.63-0.4175)/0.63 *100% = 33.7% Cu-Zn: (1.1-0.8478)/1.1*100% = 22.9%
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