Question

Need help with a, b, c

Answer 1

how is it Fe3+ when the picture says Fe2+?

Answer 2

determine the \(E_{cell}\) for both directions. from this info you can infer which is the cathode and the anode.

Answer 3

a)
Anode reaction: Al3+ + 3e- -> Al (oxidation)
Cathode reaction: Fe2+ + 2e- -> Fe (reduction)
least common number between 2 and 3 is 6... so
2(Al ---> Al3+ + 3e-) + 3(Fe2+ + 2e- ---> Fe )
2Al + 3Fe2+ + 6e- ---> Al3+ + 6e- + Fe
Overall cell reaction: 2Al + 3Fe2+ ---> Al3+ + Fe

b)
Fe^(2+)+ 2e^-→Fe E°=-0.41V
〖Al〗^(3+)+ 3e^-→Al E°=-1.66V
2Al + 3Fe2+ → Al3+ + Fe E°=1.25V

c)
Electrons mostly always flow from Anode to Cathode. In this case meaning from Aluminum (oxidizing - giving off electrons) to the Iron (reducing - receiving electrons)

this is what i got is this right or wrong?

Answer 4

it's correct.