Question

using standard potentials given in the appendices calculate the standard cell potentials and the equilibrium constants for the following reactions.

Zn(s)+( Fe^2+(aq)) ---> (Zn^2+(aq))+ Fe(s)

If anyone can show work, then I can do the rest of my homework by looking at this example.

Thank you!!!!!

Answer 1

Look up and post the reduction reaction and Eo for each half cell so we'll be using the same numbers.

Answer 2

I've checked every five minutes and no activity. So are you working on your homework?

Answer 3

Cathode (Reduction)
Half-Reaction Zn2+(aq) + 2e- -> Zn(s)

Standard Potential E° (volts)= -0.76


Cathode (Reduction)
Half-Reaction Fe2+(aq) + 2e- -> Fe(s)

Standard Potential E° (volts)= -0.41

Answer 4

Both can't be the cathode.
OK. Zn --> Zn^2+ is the half reaction in the problem. What you looked up is the standard reduction potential but this is an oxidation; therefore, reverse the sign. The E value for the half reaction as written is +0.76 v. The other is a reduction and it has the potential you have indicated. Add the oxidation half and the reduction half to obtain the total rxn.
Ecell as written is 0.76 + (-0.41 = ?

Answer 5

How do you calculate the equilibrium constants for these reactions? This was the second part of the question!

Thank you!

Answer 6

Sorry, I forgot about that.
dGo = -nFEo