What is the voltage and the Gibbs Free Energy for the following reaction? (Assume standard conditions)
3Sn^2+ (aq) + 2Al (s) --> 2 Al^3+ (aq) + 3 Sn (s)
Now calculate the Gibbs Free Energy for the above reaction if Sn^2+ (aq) is 0.15 and Al^3+ (aq) is 1.8
Look up reduction potential for Sn^2+.
Look up oxidation potential for Al.
Add the two potential for Ecell.
Using the reduction potential for Sn, plug in the values for concentration and recalculate E Sn.
For the Al, change the sign of the oxidation potential you had, plug in the concentration from the problem, calculate Ered for Al, then change the sign of that number.
Finally, add the two recalculated ox and red to find the new Ecell for the new concentrations.
To determine the voltage and the Gibbs Free Energy for a given reaction, we need to use Standard Reduction Potentials.
1. First, we need to find the half-reactions involved in the overall reaction and look up their standard reduction potentials. The half-reactions are:
Half-reaction 1: 3 Sn^2+ (aq) + 6 e- --> 3 Sn (s)
Half-reaction 2: 2 Al (s) --> 2 Al^3+ (aq) + 6 e-
2. Next, we need to find the individual standard reduction potentials for each half-reaction. The values can be found in various reference books or online databases. Let's assume the following values:
E°(half-reaction 1) = -0.150 V
E°(half-reaction 2) = -1.660 V
3. Now, we need to calculate the overall voltage (E°) of the reaction by subtracting the standard reduction potential of the anode (where oxidation occurs) from the potential of the cathode (where reduction occurs):
E°(reaction) = E°(cathode) - E°(anode)
In this case, half-reaction 1 is the cathode and half-reaction 2 is the anode. Thus:
E°(reaction) = -1.660 V - (-0.150 V)
E°(reaction) = -1.510 V
So, the voltage for the given reaction is -1.510 V.
Now let's calculate the Gibbs Free Energy (ΔG) for the reaction under the given conditions.
4. The Gibbs Free Energy change (ΔG) of a reaction is related to the standard Gibbs Free Energy change (ΔG°) through the equation:
ΔG = ΔG° + RT ln(Q)
Where:
ΔG = Gibbs Free Energy change (in J/mol)
ΔG° = Standard Gibbs Free Energy change (in J/mol)
R = Gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
Q = Reaction Quotient
5. As per the given conditions, Sn^2+ (aq) is 0.15 and Al^3+ (aq) is 1.8. These values refer to the concentrations of the ions in the reaction mixture. We will assume the temperature to be 298 K.
6. We can calculate Q (reaction quotient) as follows:
Q = [Al^3+]^2 / [Sn^2+]^3
= (1.8)^2 / (0.15)^3
≈ 230.4
7. Now we can substitute the given values into the equation to calculate ΔG:
ΔG = ΔG° + RT ln(Q)
= ΔG° + (8.314 J/(mol·K))(298 K) ln(230.4)
Here we need the ΔG° value, which is the standard Gibbs Free Energy change. If you have that value, you can substitute it in and calculate ΔG. Please note that the ΔG° value can be obtained from tables or databases.
By following these steps, you can determine the voltage and the Gibbs Free Energy for the given reaction and also calculate the Gibbs Free Energy under specific conditions.