Question

A Cu/Cu2+ concentration cell has a voltage of 0.21 V at 25 ∘C. The concentration of Cu2+ in one of the half-cells is 1.5×10−3 M .

What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell to be the lower of the two concentrations.)

Answer 1

We will use Nernst equation here, Which is

Ecell = E⁰cell - 0.0592 / n (log Q)

E⁰cell = E⁰cathode - E⁰anode

Here as the cell is made of same metal electrode (cathode and anode)

So, E⁰cathode = E⁰anode

Hence E⁰cell = 0

The equation will become

Ecell = -0.0592 / n (log Q)

n = 2

As the cell equation is

Cu⁺² + 2e-- > Cu(s)

Q = Reaction quotient = [concentration of Cu+2 in unknwon cell / Cu+2 concentration in known cell]

Q = [Cu+2] / 1.5 X 10^-3

Let us put all the values in the Nernst equation

0.21 = -0.0592/2 log ( [Cu+2] / 1.5 X 10^-3 )

-7.09 = log ( [Cu+2] / 1.5 X 10^-3 )

Taking antilog on both side

8.13 X 10^-8 = [Cu+2] / 1.5 X 10^-3

So [Cu+2] = 1.22 X 10^-10 M