A coppor, Culs). electrode is immersed in a solution that is 1,00 Mf in ammonia,

Question

A coppor, Culs). electrode is immersed in a solution that is 1,00 Mf in ammonia, NHs, and 1.00 M in teeraamminecoppert), Cu(NH a standard hydrogen electrode is used as the cathode, the cell potential, Fr, is found to be 0 084 V at 298 K ± Formation Constant from Cell Potential that is 100 M nammonsa, NH, . and 100 M netranminecoppertin Ca(NH,),I, Use the standard reduction potentials shown here to answer the questions Part A Reduction half-reactionE (V) Cur® (aq) + 2e Cu(s)| 0.337 2H' (aq) + 2e-H2 (g) | 0000 Based on the cell potential, whats te concentration of Cu in this sokusion? Express your answer with the appropriate units -? Cu- Value Units Hints My.Answers Ghve UR Review Part Incorrect: Try Again

Explanation / Answer

Let me answer the second question first, and then the first question.

The reaction 1, we can see in the reactants that we have 1+ charge, and 1:1 in mass for the reactants. In the products, we also see a 1+ charge, and a 1:1 relation of mass in the products, so this reaction is balanced.

REaction 2, we see a 1:1 relation of mass and a 1+ charge. However in the products, we also see a 1:1 mass relations but we see a 2+ charge. So this reaction is not balanced.

Reaction 3 we see that there's no charge in there, but we have in the products 3 hydrogens, while in the reactants we have only 2 hydrogen, so this reaction is also unbalanced.

Now for question 1, from the Nerst equation we solve for [Cu], the whole expression is:
E = E° - 0.059/n log [Cu(NH3)4] / [NH3][Cu]

The E° is:
E° = 0.337 - 0 = 0.337 V

0.084 = 0.337 - 0.059/2 log(1/[Cu])
0.084 = 0.337 - 0.0295log(1/[Cu])
0.084-0.337 = -0.0295log(1/Cu)
-0.253/-0.0295 = log(1/[Cu])
10(8.5762) = 1/[Cu]
[Cu] = 2.65x10-9 M

Hope this helps

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