Asked by MIKE
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?
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?
Answers
Answered by
DrBob222
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)
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)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.