Determine the equilibrium constant for the following reaction.

Question

2 Fe3+(aq) + H2(g) 2 Fe2+(aq) + 2 H+(aq)

Answer 1

Here's my work, but it wasn't the correct answer:

The two half-reactions:
Fe3+ + e- --> Fe2+ E = 0.77 V
2H+ + 2e- --> H2 E = 0 V

You multiply the first one by 2, so it looks like the formula above: 2 Fe3+(aq) + H2(g) 2 Fe2+(aq) + 2 H+(aq).

Ecell = 0.77 V

lnKsp = (2)(96485)(0.77) / (8.314)(298) = 1.1e26

I have a feeling the n (2 in the equation above) is incorrect, but I don't know why. I multiplied the half-reaction of Fe by 2.

Answer 2

Note that this is ln K and not ln Ksp.
nEF = RTlnK
Your work looks ok to me.

Answer 3

I think you should flip one of them Cause rn you have 4 e on the same side instead of then being able to cancel out