ECu = Eo - 0.0592/1[log (Cu)/(Cu^+)] = ?
EI2 = Eo - 0.0592/2[log(I^-)^2/(I2)] = ?
Determine which is the more negative voltage and reverse that half cell and add it to the other one. Change the sign on Ecell for the reversed half cell and add E values to obtain Ecell.
b.
I would write the cell reaction and use
Ecell = Eocell - 0.0592/2(log Q) where
log Q = log(products)/(reactants)
A voltaic cell is constructed that uses the following half-cell reactions.
Cu+(aq) + e− -> Cu(s)
I2(s) + 2 e− -> 2 I−(aq)
The cell is operated at 298 K with [Cu+ ] = 2.7 M and [I− ] = 2.7 M.
(a) Determine E for the cell at these concentrations.
(b) If [Cu+ ] was equal to 1.2 M, at what concentration of I− would the cell have zero potential?
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