Answer:
The concentration is [tex] [Cu^{2+}]_a = 10^{-10.269} [/tex]
Explanation:
From the question we are told that
The voltage of the cell is [tex]E = 0.22 \ V[/tex]
Generally the reaction at the cathode is
[tex]Cu^{2+} _{(aq)} + 2e^{-} \to Cu_{s}[/tex] the half cell voltage is V_c = 0.337 V
Generally the reaction at the anode is
[tex]Cu _{(s)} \to Cu^{2+} _{(aq)} + 2e^{-}[/tex] the half cell voltage is V_a = -0.337 V
Gnerally the reaction of the cell is
[tex]Cu_{(s)} + Cu^{2+} _{(aq)} \to Cu^{2+}_{(aq)} + Cu_{(s)}[/tex]
At initial the voltage is V = 0 V
Generally the voltage of the cell at 25°C is
[tex]E = V - \frac{0.0591}{n} log \frac{[Cu^{2+}] _a}{[Cu^{2+}]_c}[/tex]
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 [tex][Cu^{2+}]_c[/tex]
So
[tex]0.22= 0 - \frac{0.0591}{2} log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }[/tex]
=> [tex]-7.445 = log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }[/tex]
=> [tex]-7.445 = log [Cu^{2+}_a] - log [1.5*10^{-3}][/tex]
=> [tex]-7.445 + log [1.5*10^{-3} = log [Cu^{2+}_a] [/tex]
=> [tex]-7.445 - 2.824 = log [Cu^{2+}_a] [/tex]
Taking the antilog
=> [tex] [Cu^{2+}]_a = 10^{-10.269} [/tex]
=> [tex] [Cu^{2+}]_a = 5.38 *10^{-11} \ M [/tex]
