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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?


    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
    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^-

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