Given:
Al3+(aq) + 3e–  Al(s); E° = –1.66 V
Cl2(g) + 2e–  2Cl–(aq); E° = 1.36 V
What is ∆G° for the following cell reaction?
2AlCl3(aq)  2Al(s) + 3Cl2(g)
To calculate the standard Gibbs free energy change (∆G°) for the given cell reaction, we can use the equation:
∆G° = -nFE°
Where:
∆G° is the standard Gibbs free energy change
n is the number of moles of electrons transferred in the balanced equation
F is the Faraday constant (96,485 C/mol)
E° is the standard cell potential (also known as the standard electromotive force)
In the given reaction, the balanced equation is:
2AlCl3(aq)  2Al(s) + 3Cl2(g)
From the balanced equation, we can see that 6 moles of electrons are transferred (3 moles of electrons for each Al). Therefore, n = 6.
Now we need to calculate the standard cell potential (E°) for the given reaction. To do this, we can use the Nernst equation:
E°cell = E°cathode - E°anode
The cathode is the reduction half-reaction, Al3+(aq) + 3e–  Al(s) with E° = -1.66 V.
The anode is the oxidation half-reaction, Cl2(g) + 2e–  2Cl–(aq) with E° = 1.36 V.
Plugging these values into the equation, we get:
E°cell = 1.36 V - (-1.66 V)
E°cell = 1.36 V + 1.66 V
E°cell = 3.02 V
Now we can substitute the values of n and E° into the equation for ∆G°:
∆G° = -6 * 96,485 C/mol * 3.02 V
Calculating this, we get:
∆G° = -173,813 J/mol
Therefore, the standard Gibbs free energy change (∆G°) for the given cell reaction is -173,813 J/mol.