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

Answer:

The concentration is  $[Cu^{2+}]_a = 10^{-10.269}$  

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  

       $E = V - \frac{0.0591}{n} log \frac{[Cu^{2+}] _a}{[Cu^{2+}]_c}$

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

      $0.22= 0 - \frac{0.0591}{2} log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }$

=>    $-7.445 = log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }$

=>  $-7.445 = log [Cu^{2+}_a] - log [1.5*10^{-3}]$

=>    $-7.445 + log [1.5*10^{-3} = log [Cu^{2+}_a]$

=>    $-7.445 - 2.824 = log [Cu^{2+}_a]$

Taking the antilog

=>    $[Cu^{2+}]_a = 10^{-10.269}$    

=>     $[Cu^{2+}]_a = 5.38 *10^{-11} \ M$