Question

A student measures the potential of a cell constructed by immersing a copper strip into a l M CuSO4 solution and a silver wire immersed in a l M AgNO3 solution. The two solutions are connected by a salt bridge. She measures a potential of 0.45 V, with the Cu electrode being the negative electrode. 7. The student adds 6 M NH3 to be CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)4 2+ ion. The potential of the cell, Ecell, goes up to 0.92 V and the Cu electrode is still negative. What is the concentration of the Cu2+ ion under these conditions?

Answer 1

Solution:

First of all we assume that the temperature is 25°C.

In a voltaic cell anode is the negative electrode where oxidation takes place and cathode is the positive electrode where reduction takes place. We are given that copper is the negative electrode and hence is the anode.

The cell reactions can be written as

Oxidation:   Cu(aq) → Cu²⁺ (aq)   + 2e                 E° (Oxidation) = -0.34 V

Reduction: : 2Ag⁺  + 2 e → 2Ag(aq)            E° (Reduction) = + 0.80 V

.......................................................................................................................

Net reaction: Cu(aq)   +   2 Ag⁺ (aq)   → Cu²⁺ (aq) + 2Ag        E° (Cell) = 0.46 V

After the student adds 6M ammonia to the copper sulfate solution, producing the complex ion Cu(NH3)₄²⁺ (aq).Because the cell is not operating under standard conditions, the Nernst equation would need to be used. The formula for the Nernst equation is

E(cell) = E°(cell) - (0.0592/n) log Q at 25°C …………………(1)

Where Q = ([Cu^2+]/[Ag^+]^2)

                 = x/ (1)^2                 (Let x be the concentration of Cu²⁺ after addition of ammonia)  

                  = x   

(since adding NH3 into the CuSO4 compartment does not influence [Ag+] in the Ag+ compartment.so [Ag+ = 1 M, given)

Substituting for E(cell) = 0.92 V and E° (cell) = 0.45    (both values are given but according to calculation E° (cell) should be 0.46 V, we are using the given value of 0.45 V) we get

0.92 = 0.45 - (0.0592/2)logx

Or 0.92 – 0.45 = - 0.0296logx

Or 0.47/.0296 = - logx

15.88 = -logx

Or x = antilog(-15.88) = 1.32 x 10^-16

Therefore the concentration of Cu²⁺ under the given conditions is 1.32 x 10^-16 M