The process of using an electric current to bring about chemical change is called electrolysis.Electrolysis is a processes of oxidation and reduction at the respective electrodes due to external current passed in the electrolyte. The product obtained during electrolysis depends on following factors. The nature of the electrolyte The concentration of electrolyte The nature of the electrode. Consider the electrolysis of following cell containing aq. solution of CuSO_4,ZnCl_2 and MBr_2 by using pure silver rod as a cathode and Pt electrode as anode.Assume that M^(2+) does not further oxidise and can not form complex with NH_3 Assume no hydrolysis of any ion. E_(Cu^(2+)//Cu)^(0)=0.35 V, E_(M^(2+)//M)^(0)=-0.10 V, E_(Zn^(2+)//Zn)^(0)=-0.76 V, E_(H_(2)O // H_2)^(0)=-0.828 V, E_(Ag^(+)//Ag)^(0)=0.80 V, (2.303RT)/F=0.06 After passing 20 amp current from battery for 28950 sec.the remaining conc. of ions in solution given in passage would be: (Assume current efficiency to be 100%)

The process of using an electric current to bring about chemical chang

1.	The process of using an electric current to bring about chemical change is called electrolysis.Electrolysis is a processes of oxidation and reduction at the respective electrodes due to external current passed in the electrolyte. The product obtained during electrolysis depends on following factors. The nature of the electrolyte The concentration of electrolyte The nature of the electrode. Consider the electrolysis of following cell containing aq. solution of CuSO_4,ZnCl_2 and MBr_2 by using pure silver rod as a cathode and Pt electrode as anode.Assume that M^(2+) does not further oxidise and can not form complex with NH_3 Assume no hydrolysis of any ion. E_(Cu^(2+)//Cu)^(0)=0.35 V, E_(M^(2+)//M)^(0)=-0.10 V, E_(Zn^(2+)//Zn)^(0)=-0.76 V, E_(H_(2)O // H_2)^(0)=-0.828 V, E_(Ag^(+)//Ag)^(0)=0.80 V, (2.303RT)/F=0.06 After passing 20 amp current from battery for 28950 sec.the remaining conc. of ions in solution given in passage would be: (Assume current efficiency to be 100%)
Answer» `[CU^(2+)]=0.5M, [M^(2+)]=0.5M, [Zn^(2+)]=0.1M` `[Cu^(2+)]~~0M, [M^(2+)]=0.5M, [Zn^(2+)]=0.5M` `[Cu^(2+)]=0.5M, [M^(2+)]=0.5M, [Zn^(2+)]=0.5M` `[Cu^(2+)]~~0M, [M^(2+)]=0.5M, [Zn^(2+)]=1 M` SOLUTION :Deposition ORDER `CugtMgtZn` Cu required `implies (Ixxt)/96500=w/E` `20xxt=1xx2xx2xx96500` t=19300 SEC. In 28950-19300=9650 sec `M^(2+)`, will deposit as M. mole of `Mxx2=(20xx9650)/96500` mole of M deposit =1 Initial mole of `M^(2+)=1xx2`,remain `M^(2+)` moles =2-1=1 ,`[M^(2+)]=1/2=0.5`