Question

Using these metal ion/metal standard reduction potentials calculate cell potential for Cu2+(aq) + Cd(s) →Cd2+(aq)+ Cu(s) Cu2+(aq)|Cu(s) + 0.34 VNi2+(aq)|Ni(s) -0.25 Cd2+(aq)/Cd(s) -0.4V

Answer 1

Solution :

Cd(s) ---------------------->  $Cd^{+2}$ (aq) + 2$e^-$      ,     $E_0$  =   0.34 v

$Cu^{+2}$  (aq)  +   2$e^-$  ------------> Cu (s)             ,     $E_0$  =  -0.04 v

----------------------------------------------------------------------------------------------

Cd(s)  +  $Cu^{+2}$  (aq)  ------------->  $Cd^{+2}$ (aq)  +  Cu (s)   ,      $E_0$  =  0.30 v  

The cell potential is defined as the measure of $\text{ potential difference }$  between the $\text{two half cells}$ of an electrochemical cell.