Question

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?

Answer 1

2Al(s) + The vol3Pb^+2(aq) ==> 2Al^+3 3Pb(s)
The voltage is1.66 + (reduction potential for Pb+2 ==> Pb(s))

Answer 2

I receive the answer: -442866.15 J or -443 kJ with significant figures

Answer 3

the voltage would be 1.66 + -.13 which is 1.53V

Answer 4

hkae