**Consider the galvanic cell based on the following half-reactions: Au^3+(aq) + 3 e^-→ Au (s) E° = 1.50 V Tl^+(aq) + e^-→ Tl (s) E° = -0.34 V a) Determine the overall cell reaction and calculate E°cell b) Calculate delta G° and the equilibrium constant K for the cell reaction at 25°C**

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

**Consider the galvanic cell based on the following half-reactions: Au^3+(aq) + 3 e^-→ Au (s) E° = 1.50 V Tl^+(aq) + e^-→ Tl (s) E° = -0.34 Va) Determine the overall cell reaction and calculate E°cell b) Calculate delta G° and the equilibrium constant K for the cell reaction at 25°C c) Calculate Ecell at 25°C when [Au3+] = 1.0 x 10^-2M and [Tl+] = 1.0 x 10^-4M. Will the cell potential increase or decrease by increasing the concentration of Au3+? **

For parts A & B, see http://www.enotes.com/chemistry/q-and-a/consider-galvanic-cell-based-following-half-333659

Please note that when you post a question, it should only include *one* question. Dividing it into parts A-C does not make it one question :)

For part C, we need to calculate Ecell under non-standard conditions

Ecell = Eocell - 0.0592/n log Q

We know Eocell and have 1.84V (see link above). To find Q, we have to first write the balanced equation

Au3+(aq) + 3Tl(s) --> Au(s) + 3Tl+(aq)

We need the 3s in front of Tl and Tl+ so that the number of electrons in each reaction is the same so that they will cancel out. Now, we can find Q

Q = [Tl+]^3 / [Au3+]

Q = (1x10^-4)^3 / 1x10^-2

Q = 1 x 10^-10

Now we can solve for Ecell

Ecell = 1.84 - (0.0592/3) log (1 x 10^-10)

Ecell = 2.04 V

If we increase the concentration of Au3+, then the value of Q will decrease and the log of the value will also decrease. As a result, the second term will have a greater magnitude so the Ecell will increase because we are substracting a negative term.

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