If we are looking at a galvanic cell, then it be a spontaneous process so we need to determine which is the oxidation reaction and which one is the reduction reaction. The half-reaction with the more positive reduction potential will be reduced and the half-reaction with the more negative reduction potential will be oxidized.

The gold will be the reduction reaction and the thallium will be the oxidized reaction. To find Eocell, we use the reduction potentials and take Ecathode - E anode

Thereore, we will get

1.50 V - (-0.34 V) = 1.84 V

Since Eocell is positive, this agrees with the fact that this is a galvanic cell.

To find deltaGo, we can use

deltaG = - nFEocell

where n is the number of electrons transferred (3 in this case) and F is Faraday's constant, 96485 C/mole of e-.

deltaGo = -(3)(96485)(1.84)

deltaGo = 533 kJ/mol

To find K, we can look at the relationship

Eocell = 0.0592/n log K

1.84 = 0.0592/3 log K

log K = 93.2

K = 1.8 x 10 ^93

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