Given:
Al3+(aq) + 3e– --> Al(s); E° = –1.66 V
I2(s) + 2e– --> 2I–(aq); E° = 0.54 V
What is ∆G° for the following cell reaction?
2AlI3(aq) --> 2Al(s) + 3I2(s)
A) 4.2 × 105 J
B) –6.5 × 105 J
C) –1.3 × 106 J
D) 1.3 × 106 J
E)
–4.2 × 105 J
To find ΔG° for the cell reaction, we can use the relationship between ΔG° and the standard cell potential, E°:
ΔG° = -nFΔE°
where ΔE° is the difference in standard cell potentials between the reduction half-reaction and the oxidation half-reaction, n is the number of moles of electrons transferred in the balanced chemical equation, and F is the Faraday constant (96,485 C/mol).
In this case, the reduction half-reaction is:
I2(s) + 2e– --> 2I–(aq); E° = 0.54 V
and the oxidation half-reaction is:
Al3+(aq) + 3e– --> Al(s); E° = –1.66 V
The overall cell potential can be found by subtracting the oxidation potential from the reduction potential:
E°cell = E°reduction - E°oxidation
E°cell = 0.54 V - (-1.66 V)
E°cell = 2.20 V
The number of moles of electrons transferred can be determined from the balanced chemical equation for the overall cell reaction:
2AlI3(aq) --> 2Al(s) + 3I2(s)
Since 2 moles of aluminum are produced for every 2 moles of AlI3 consumed, and 3 moles of iodine are produced for every 2 moles of AlI3 consumed, the overall number of moles of electrons transferred is 2.
Plugging in the values into the equation for ΔG°:
ΔG° = -nFΔE°
ΔG° = -(2)(96,485 C/mol)(2.20 V)
ΔG° = -4.18 × 10^5 J
Therefore, the correct answer is E) –4.2 × 10^5 J.
To find ∆G° for the given cell reaction, we can use the equation:
∆G° = -nF∆E°
Where:
∆G° = standard Gibbs free energy change
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant (96,485 C/mol)
∆E° = standard cell potential difference
First, let's determine the number of moles of electrons transferred in the balanced equation. By examining the balanced equation:
2AlI3(aq) --> 2Al(s) + 3I2(s)
We can see that 6 moles of electrons are transferred from the left-hand side to the right-hand side.
Next, we can calculate the overall standard cell potential difference (∆E°) by summing the individual standard cell potentials:
∆E° = ∆E°(cathode) - ∆E°(anode)
∆E°(cathode) = ∆E° for reduction half-reaction (I2(s) + 2e– --> 2I–(aq)) = 0.54 V
∆E°(anode) = –∆E° for oxidation half-reaction (Al3+(aq) + 3e– --> Al(s)) = –(–1.66 V) = 1.66 V
∆E° = 0.54 V - 1.66 V = -1.12 V
Finally, we can substitute the values into the equation to find ∆G°:
∆G° = -nF∆E°
∆G° = -(6 mol)(96,485 C/mol)(-1.12 V)
∆G° = 6 × 96,485 × 1.12 J
∆G° ≈ 6.5 × 105 J
Therefore, the correct answer is option B) –6.5 × 105 J.
2Al^3+ + 6I^- ==> 2Al(s) + 3I2(s)
dGo = -nFEo
n = 6