When you post problems that depend upon numbers you would do well to post the numbers in your text because texts change over the years and those constants change. The E value I looked up for I is =0.535 and the E value for Cu is +0.521.
Step 1 is to use the reduction ernst equation, substitute 2.7M and calculate E for I. I get +.509 but your values may be different. Do the same for Cu and I obtained 0.469; again your numbers may be different.
Then write the equation.
I2 ==> 2I^- E = +0.509v
Cu + e ==> Cu E = -0.469v
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I2 + 2Cu ==> 2Cu^+ + 2I^- Ecell = sum of E values.
To do b I would use the overall equation.
Ecell = EoCell - (0.0592/2)logQ
log Q = (I^-)^2(Cu^+)^2/((Cu)(I2) and solve for I^-
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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