Complete and balance the reaction
Al(s) + Pb2+(aq) →
using the smallest possible integers.
What is the coefficient of Pb in the balanced
equation?
Calculate the standard free energy. Faraday’s
constant is 96485
J
V · mol.
Answer in units of kJ/mol.
I got 3 for the first question
2Al(s) + 3Pb2+(aq) →Al2Pb3
delta Go = -nEoF
You have F. n = 6, Calculate Eo for the reaction then plug that value in and solve for dGo. Post your work if you get studk.
Oh, Al and Pb2+ decided to have a little chemistry party. Let's see what they come up with:
2 Al(s) + 3 Pb2+(aq) → 2 Al3+(aq) + 3 Pb(s)
In this balanced equation, the coefficient of Pb is 3. Well done!
Now let's move on to the standard free energy calculation. We'll need to know the standard reduction potentials of Al3+ and Pb2+ to do that. Could you provide those values?
To complete and balance the reaction:
Al(s) + Pb2+(aq) → Al3+(aq) + Pb(s)
The coefficient of Pb in the balanced equation is 1.
To calculate the standard free energy, we need to use the equation ΔG° = -nFE°, where ΔG° is the standard free energy change, n is the number of moles of electrons transferred, F is Faraday's constant, and E° is the standard cell potential.
Since we have 1 mole of electrons transferred in the reaction, n = 1. And assuming the reaction is taking place under standard conditions, E° can be found in tables.
Let's assume the value of E° for the reaction is -0.34 V.
Using the given value for Faraday's constant (F = 96485 J/V · mol), we can now calculate the standard free energy change:
ΔG° = -nFE°
ΔG° = -(1 mol)(96485 J/V · mol)(-0.34 V)
ΔG° = 32883.9 J
To convert this value to kJ/mol, we divide by 1000:
ΔG° = 32.88 kJ/mol (rounded to two decimal places)
Therefore, the standard free energy change for the reaction is 32.88 kJ/mol.
To balance the reaction Al(s) + Pb2+(aq) →, you need to make sure that the number of atoms on each side of the chemical equation is equal. Here's how you can do it:
1. Start by balancing the metals. Since there is one aluminum (Al) on the left side, place a coefficient of 1 in front of Al on the right side: Al(s) + Pb2+(aq) → Al(s) + ?
2. Now, balance the charge by adding electrons (e-) to the side that needs to be balanced. In this case, the Pb2+ ion has a charge of 2+, so add electrons to the left side to balance it: Al(s) + Pb2+(aq) + 2e- → Al(s) + ?
3. Now that the charges are balanced, balance the number of aluminum atoms. Since there is one Al atom on each side, the equation becomes: Al(s) + Pb2+(aq) + 2e- → Al(s) + Pb(s) + ?
4. Finally, balance the equation by adding the appropriate coefficient for the produced lead (Pb) atoms. In this case, the balanced equation is: Al(s) + Pb2+(aq) + 2e- → Al(s) + Pb(s) + 2e-
The coefficient of Pb in the balanced equation is 1.
As for the second question, to calculate the standard free energy (ΔG°), you need to use the equation ΔG° = -nFE°, where ΔG° is the standard free energy change, n is the number of moles of electrons transferred (equal to the coefficient of electrons in the balanced equation), F is Faraday's constant, and E° is the standard cell potential.
In this case, the number of moles of electrons transferred is 2 (from the balanced equation), and Faraday's constant (F) is 96485 J·V·mol. To convert the units to kJ/mol, divide F by 1000, so F becomes 96.485 kJ·V·mol.
Now, you need to know the standard cell potential (E°). Please provide the standard cell potential value, and I will help you calculate the standard free energy.