Calculate the E° for each of the following reaction. Decide whether each is product-favored in the direction written

2 Cl-(aq) + Cu2+(aq) --> Cu(s) + Cl2(g)

i got -1.020 but its wrong and i do not know why.

If you will show the numbers you looked up and how you worked the problem I may be able to find the error. I looked the numbers up in an old old book of mine and found -1.02 v and the reaction is favored toward the reactants (not the products). I suspect the problem, and apparently you are keying the numbers into a computer database, is the last zero. The correct answer may be -1.02 and NOT -1.020 but I can't be certain of that without seeing your solution. What I'm saying is that I think you have included too many significant figures.

To calculate the standard cell potential (E°) for a reaction, you need to use the standard reduction potentials (E°) for each half-reaction involved and apply the Nernst equation.

In this case, we have the following half-reactions:
1. Cu2+(aq) + 2e- --> Cu(s) (reduction half-reaction)
2. Cl2(g) + 2e- --> 2 Cl-(aq) (oxidation half-reaction)

The standard reduction potentials (E°) for these reactions are given in a standard reduction potentials table. Let's assume the E° for the reduction of Cu2+ to Cu is +0.34 V, and the E° for the reduction of Cl2 to Cl- is +1.36 V.

Since these half-reactions are written as reductions, you need to flip the sign of the E° values for the oxidation half-reactions.

Step 1: Calculate the overall cell potential (E°cell) by subtracting the E° of the oxidation half-reaction from the E° of the reduction half-reaction:
E°cell = E°reduction - E°oxidation
E°cell = (+0.34 V) - (-1.36 V)
E°cell = +1.70 V

Step 2: Determine the direction of the reaction based on the sign of E°cell. A positive E°cell indicates that the reaction is spontaneous in the forward direction (product-favored). If E°cell is negative, the reaction is non-spontaneous in the forward direction (reactant-favored).

In this case, since the E°cell is +1.70 V, the reaction is product-favored in the direction as written.

Therefore, the correct value for E° for the given reaction is +1.70 V, not -1.020 V. Make sure to double-check the standard reduction potentials used and their signs in your calculations to avoid errors.