Using a table of standard electrode potentials (as in Appendix M of your text), calculate the standard cell potential, Eo, for: 2 Fe2+(aq) + Cl2(g) 2 Fe3+(aq) + 2 Cl(aq).

Answer:

Anode: 2 Fe2+ 2 Fe3+ + 2 e
Eox = 0.771 V
Cathode: Cl2 + 2 e 2 Cl Ered = +1.36 V
Enet = +0.59 V

I think Eox = -0.771 and

Ered = +1.36
Then 1.36 + (-0.771) = ?
Your answer is right; I don't agree with what you called Eox. The anode is oxidized, yes, and that is an oxidation, yes. But the potential for that electrode is -0.771. The reduction potential is 0.771; the oxidn potential is -0.771.

To calculate the standard cell potential, Eo, for the reaction given, you need to find the standard electrode potentials (Eo) for each half-reaction, and then subtract the reduction potential from the oxidation potential.

First, locate the half-reactions in the table of standard electrode potentials. The oxidation half-reaction is 2 Fe2+(aq) 2 Fe3+(aq) + 2 e- and the reduction half-reaction is Cl2(g) + 2 e- 2 Cl-(aq).

Next, find the standard electrode potentials (Eo) for each half-reaction. The Eo value for the oxidation half-reaction, Eox, is given as 0.771 V, and the Eo value for the reduction half-reaction, Ered, is given as +1.36 V.

To calculate the overall standard cell potential, subtract the reduction potential (Ered) from the oxidation potential (Eox):

Eo = Eox - Ered
Eo = 0.771 V - (+1.36 V)
Eo = 0.771 V - 1.36 V
Eo = -0.59 V

Therefore, the standard cell potential, Eo, for the given reaction is -0.59 V.