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 If 36 mol of NH_(3)(g) is passed in electrolytic solution given in comprehension (assume no volume change by addition of NH_(3)),then what would be decreasing order of reduction potential of following :

Answer»

`M^(2+)gtCu^(2+)gtZn^(2+)`
`CU^(2+)gtM^(2+)gtZn^(2+)`
`M^(2+)gtZn^(2+)gtCu^(2+)`
`Cu^(2+)gtM^(2+)gtZn^(2+)`

Solution :`{:(Zn^(2+)+ 4NH_3 hArr Zn(NH_3)_4^(2+)" " K_f=1xx10^(9)),(1 "" 18),(1-x " " 18-4x-4y " " x):}`
`Zn^(2+)=(Zn(NH_3)_4^(2+))/(K_f[NH_3]^4)=1/(10^9xx10^4)=10^(-)13`
For `Zn^(2+)+2e^(-)toZn(s)`
`E_(R.P)=-0.76-0.06/2"log"1/10^(-13)=-0.76-0.06xx13/2=-1.15 V`
`{:(Cu^(2+)+ 4NH_3 hArr Cu(NH_3)_4^(2+)" " K_f=1xx10^(12)),(1 "" 18),(1-x " " 18-4x-4y " " y):}`
`[Cu^(2+)]=(Cu(NH_3)_4^(2+))/(K_f[NH_3]^(4))=10^(-16)`
For `Cu^(2+)+2e^(-)TOCU(s)`
`E_(R.P)=-0.34-0.06/2"log"1/10^(-16)=-0.34-0.06xx8=-1.14 V`


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