1.

Zinc rod is dipped in 0.1 M solution of ZnSO_4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given, E_(Zn^(2+)//Zn)^(@) = -0.76 V]

Answer»

Solution :Applying Nernst Equation
`E_(Zn^(2+)//Zn) = E_(Zn^(2+)//Zn)^(@) -0.0591/2 LOG 1/([Zn^(2+)])`
`[Zn^(2+)] =0.1 XX 95/100= 0.095 M`
`Zn^(2+) + 2e^(-) to Zn` or n=2
`E_(Zn^(2+)//Zn) = -0.76 V- 0.0591/2 [log1000- log 95]=-0.76 V -(0.0591)/2 [ 3.000 - 1.9777]`
`=-0.76 V - (0.0591)/2 xx 1.0223 = -0.76 V - 0.00604/2`
`=-0.76 V - 0.0302 = -0.7902` V
Electrode potential `=-0.7902` V


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