1.

Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+) E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(o) is

Answer»

`-0.10V`
`+0.18V`
`-0.54V`
`0.54V`

Solution :`I^(-)` get OXIDIZED to `I_(2)` hence will FORM anode and `Cr_(2)O_(7)^(2-)` get REDUCED to `Cr^(3+)` hence will form CATHODE.
`E_(cell)^(o)=E_("cathode")^(o)-E_("anode")^(o),E_(cell)^(o)=E_(Cr_(2)O_(7)^(-2))-E_(I_(2))^(o)`.
`0.79=1.33-E_(I_(2))^(o),E_(I_(2))^(o)=1.33-0.79,E_(I_(2))^(o)=0.54V`.


Discussion

No Comment Found

Related InterviewSolutions