Question

A copper, Cu(s), electrode is immersed in a solution that is 1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II), [Cu(NH3)4]^{2+}. If a standard hydrogen electrode is used as the cathode, the cell potential, Ecell, is found to be 0.075 V at 298 K.
Half reaction of Cu = .337 V
Half reaction of H = .000 V

Based on the cell potential, what is the concentration of Cu^{2+} in this solution?

[Cu2+]____ with units


PLEASE HELP ME!!

Answer 1

2H⁺(aq) + Cu(s) + 4NH₃(aq) --> H₂(g) + [Cu(NH₃)₄]²⁺

Ecell = E^o - RT/nF (ln Q)

Q = [Cu(NH3)4]²⁺/ [NH3]⁴ = 1 M / 1M = 1

Q = 1

Ecell = E^o = 0.075 V

This process(reference) can be written as addition of 3 steps ( I + II + III) :

I. 2H⁺(aq) + 2e^- --> H₂(g)

II. Cu(s) --> Cu²⁺(aq) + 2e^-

III. Cu²⁺(aq) + 4NH₃(aq) --> [Cu(NH₃)₄]²⁺(aq)

E_ref = E_I + E_II + E_III

E_ref is known from the experiment

E_ref = 0.075 V

E_I is 0.00 V ( reduction potential of standard hydrogen electrode = 0.0 V )

E_II = - ( standard reduction potential of Cu(II) )

E_II = - 0.337 V

E_III = 0.075 - ( 0.0 - 0.337) V

E_III = 0.075 + 0.337 = 0.412 V

Now,

E^o = 0.0591/n * log ( [Cu(NH3)4]²⁺ / [Cu²⁺] * [NH3]^4)

0.412 V = 0.0591 / 2 * log ( 1/[Cu²⁺] )                 Since [Cu(NH3)4]²⁺ = [NH3] = 1 M and n = 2 because 2e- are required to reduce Cu2+ to Cu

0.412 * 2 / 0.0591 = log ( 1/[Cu²⁺] )

13.942 = log ( 1/[Cu²⁺] )

1/[Cu²⁺] = 10^13.942 = 8.76 x 10^13

[Cu²⁺] = 1.142 x 10^-14 M