1.

Zn^(2+) to Zn_((s)),E^(@)=-0.76V Cu^(2+)toCu_((s)),E^(@)=-0.34V which of the following is spontaneous

Answer»

`ZN^(2+)toCutoZn+Cu^(2+)`
`Cu^(2+)+ZntoCu+Zn^(2+)`
`Zn^(2+)+Cu^(2+)toZn+Cu`
none of these

Solution :`Zn^(2+)toZn_((s)),E^(o)=-0.76V`
`Cu^(2+)toCu_((s)),E^(o)=-0.34V`
A redox reaction is FEASIBLE if `E_(cell)` is positive.
`Zn^(2+)+2e^(-)toZn_((s)),E^(o)=-0.76V`
`UNDERLINE(Cu^(2+)+2e^(-)toCu_((s)),E^(o)=-0.34V"")`
`thereforeCu^(2+)+ZntoZn^(2+)+Cu,E_(cell)=E_((Cu^(2+)//Cu))^(o)-E_((Zn^(2+)//Zn))^(o)`
Therefore, `E_(cell)=-0.34-(-0.76)`
`=-0.34+0.76=+0.42V`


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