1.

For the cell reaction, 2Cr^(4+)+Coto2Ce^(3+)+Co^(2+)""E_((cell))^(@) is 1.89V and E_(Co//Co^(2+))^(@)=-0.028. If E_(Ce^(4+)//Ce^(3+))^(@)

Answer»

`-1.64V`
`+1.64V`
`-2.08V`
`+2.17V`

SOLUTION :In this cell Co is oxidised and it ACTS as anode and Ce acts as cathode.
`E_(Cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=1.89=E_(Cell)^(@)-(-0.28)`
`E_(Cell)^(@)=-1.89-0.28=1.61`VOLT.


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