Question

A Cu/Cu2 concentration cell has a voltage of 0.22 V at 25 o C. The concentration of Cu2 in one of the half-cells is 1.5 x 10-3 M. What is the concentration of Cu2 in the other half-cell

Answer 1

The concentration is
`[Cu^(2+)]_a  = 10^(-10.269)`

Step-by-step explanation:

From the question we are told that

The voltage of the cell is
`E =  0.22 \  V`

Generally the reaction at the cathode is

`Cu^(2+) _((aq)) + 2e^(-) \to  Cu_(s)` the half cell voltage is V_c = 0.337 V

Generally the reaction at the anode is

`Cu _((s)) \to Cu^(2+) _((aq)) +  2e^(-)` the half cell voltage is V_a = -0.337 V

Gnerally the reaction of the cell is

`Cu_((s)) + Cu^(2+) _((aq)) \to Cu^(2+)_((aq)) +  Cu_((s))`

At initial the voltage is V = 0 V

Generally the voltage of the cell at 25°C is

`E =  V  - (0.0591)/(n) log ([Cu^(2+)] _a)/([Cu^(2+)]_c)`

Here n is number of of electron and it is 2

So from the question we are told that one cell has a concentration 1.5 x 10-3 M

Let assume it is
`[Cu^(2+)]_c`

So

`0.22=  0  - (0.0591)/(2) log ([Cu^(2+)] _a)/(  1.5 * 10^(-3) )`

=>
`-7.445 =     log ([Cu^(2+)] _a)/(  1.5 * 10^(-3) )`

=>
`-7.445 =     log [Cu^(2+)_a] - log [1.5*10^(-3)]`

=>
`-7.445  + log [1.5*10^(-3) =     log [Cu^(2+)_a]`

=>
`-7.445  - 2.824 =     log [Cu^(2+)_a]`

Taking the antilog

=>
`[Cu^(2+)]_a  = 10^(-10.269)`

=>
`[Cu^(2+)]_a  = 5.38 *10^(-11) \  M`