Question

Use the standard reduction potentials shown here to answer the questions.

Reduction half-reaction	E∘ (V)
Cu2+(aq)+2e−→Cu(s)	0.337
2H+(aq)+2e−→H2(g)	0.000

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 0.073 V at 298 K.

Part A

Based on the cell potential, what is the concentration of Cu2+ in this solution?

Express your answer to one significant figure and include the appropriate units.

Answer 1

Dear student,

The H electrode to form the anode and the standard E(cell) will be +0.337 v

now assume the [Cu(NH₃)₄]²⁺ is not directly involved in the measured cell potential only Cu²⁺(aq) is.

Then use the Nernst equation to find [Cu²⁺⁽aq)]

E(cell) = standard E(cell) - (0.0592/n).log Q ---------------- Formula

where in this case Q = 1/[Cu²⁺(aq)] and n = 2

E(cell)=0.073 V, standard E(cell)=0.337 V, n=2 calculate Cu²⁺=?

Put the values in formula

0.073 = 0.337 –(0.0592/2) log(1/ Cu²⁺)

0.073 + (0.0296) log(1/ Cu²⁺) = 0.337

(0.0296) log(1/ Cu²⁺)=0.264

log(1/ Cu²⁺)= 0.264/0.0296

log(1/ Cu²⁺)=8.92

(1/Cu2+) = 10^8.92

Cu²⁺=1.20 x 10^-9 M