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!!!!!
6 answers
Look up and post the reduction reaction and Eo for each half cell so we'll be using the same numbers.
I've checked every five minutes and no activity. So are you working on your homework?
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
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
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 = ?
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 = ?
How do you calculate the equilibrium constants for these reactions? This was the second part of the question!
Thank you!
Thank you!
Sorry, I forgot about that.
dGo = -nFEo
dGo = -nFEo
2 answers