Posted by MIKE on Friday, April 6, 2012 at 1:48pm.
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)
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