Complete and balance the reaction
Al(s) + Pb2+(aq) !
using the smallest possible integers.
I got:
2Al(s) + 3Pb2 ==> Al2Pb3
Calculate the standard free energy. Faraday’s
constant is 96485
J V · mol
Answer in units of kJ/mol.
I know the formula is -nFE but when I plug in the numbers
-(3)(96485)(-1.66V)
I get the wrong answer
Can you please tell me what I'm doing wrong?
the voltage would be 1.66 + -.13 which is 1.53V
Well, it looks like you're on the right track with the formula -nFE for calculating the standard free energy change (ΔG). However, I believe the issue lies in the value you're using for the voltage (E).
The voltage should represent the cell potential or the difference in electrode potentials between the half-reactions involved in the redox reaction. Since you haven't provided the half-reactions, I cannot determine the correct voltage value.
Once you have the correct voltage value, you can proceed with the calculation of ΔG using the formula -nFE.
Remember, ΔG is expressed in joules, so to convert it to kJ/mol, you need to divide the result by 1000.
Hope that helps!
To complete and balance the reaction, Al(s) + Pb2+(aq) should give Al3+(aq) + Pb(s). The balanced equation would be:
2Al(s) + 3Pb2+(aq) -> 2Al3+(aq) + 3Pb(s)
Now let's calculate the standard free energy using the formula ΔG = -nFE, where ΔG is the standard free energy change, n is the number of moles of electrons transferred, F is Faraday's constant (96485 J V•mol), and E is the cell potential.
In this case, n is the number of electrons transferred, which is equal to 3 since each Pb2+ ion gains 2 electrons and each Al3+ ion loses 3 electrons.
Now, let's calculate ΔG:
ΔG = -(3 mol)(96485 J V•mol)(-1.66 V)
Notice that E should have a positive value since it represents the potential of the reduction half-reaction (Pb2+ to Pb(s)).
Therefore, the correct calculation should be:
ΔG = -(3 mol)(96485 J V•mol)(1.66 V)
Make sure to use the positive value for E, and the result will be the correct answer in J. To convert it to kJ/mol, divide the result by 1000:
ΔG = (3 mol)(96485 J V•mol)(1.66 V) / 1000 = 478.3 kJ/mol
So, the standard free energy change for this reaction is 478.3 kJ/mol.
2Al(s) + The vol3Pb^+2(aq) ==> 2Al^+3 3Pb(s)
The voltage is1.66 + (reduction potential for Pb+2 ==> Pb(s))