Calculate both the cell potential (voltage) and the Gibbs Free Energy generated by the cell which contains the above reaction, assuming standard conditions.
Cr2O7(-2) + I(-1) --> Cr (+3) + I2 (s)
Reduction: Cr2O7(-2) + 6e- -> 2Cr(+3) ; E = 1.33 V
Oxidation: 2I(-1) -> I2 + 2e- ; E = 0.54 V
Ecell = Ecathode - Eanode = 1.33 - 0.54 = 0.79 V
deltaG = -nFE = -(6)(96.485)(0.79) = -457 J/mol
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