Answer to Question #98941 in General Chemistry for Ephraim

Question

During the electrolysis of CU (2) tetraoxosulphate (6) solution using platinum electrode,0.816g of Cu was deposited on the cathode.calculate the volume of oxygen gas liberated at S.T.P

Answer 1

This process can be expressed with the following equation:

$2CuSO_4+2H_2O \rightarrow 2Cu \downarrow+2H_2SO_4+O_2 \uparrow;$

Now, we have to found the amount of copper metal formed on the cathode in moles:

$n(Cu)=\frac{m(Cu)}{M(Cu)}=\frac{0.816g}{63.5g/mol}=0.0129mol;$

According to the equation, the amount of oxygen formed in moles is half of that of the copper metal:

$n(O_2)=\frac{1}{2}*n(Cu)=\frac{1}{2}*0.0129mol=6.45*10^{-3}mol;$

Finally, the volume of the released oxygen under STP is:

$V(O_2)=n(O_2)*V_m=6.45*10^{-3}mol*22.4L=0.144L$ or 144 mL;

Thus, it was 144 mL of oxygen released in the electrolysis.