1.

A zinc rod is placed in 0.095 M zinc chloride solution at 25^(@)C. EmF of this half cell is -0.79V. Calculate E^(@)""_(Zn2+//Zn).

Answer»

Solution :Given : E = -0.79 V
n = 2 `[ZN^(+)]=0.095 M`
Formula : `E=E^(@)-0.0591/n log Zn^(2+)`
Solution :
`E^(0)""_(Zn^(2+)//Zn)=E+0.0591/2logZn^(2+)`
`E^(0)""_(Zn^(2+)//Zn)=-0.76+0.0591/2log0.095`
`""=-0.79+0.02889=-0.76V`
`""E^(@)=-0.76V`.


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