1.

Under which of the following conditions E value of the cell, for the cell reaction given is maximum? Zn(s) +Cu^(2+) (aq) hArrCu(s)+Zn^(2+) (aq)

Answer»

`C_1=0.1M,C_2=0.01M`
`C_1=0.01 M, C_2=0.1 M`
`C_1=0.1 M, C_2=0.2 M`
`C_1=0.2M,C_2=0.1 M`

Solution :From Nernst equation
`E=E^@-(2.303RT)/(nF)logQ`
`impliesE= E^@-(2.30 3RT)/(nF)log ((ZN^(2+))/(Cu^(2+)))`
`E_(CELL)^@=E_c^@-E_A^@=0.34-(-0.76)V=1.1V`
`E=1.1-0.059/nlog""C_2/C_1(Zn^(2+)=C_2 Cu^(2+)=C_1)`
By analysing from the above equation, E VALUE of the cell will be maximum when , `log""C_2/C_1` WOULD come out to be minimum, when `log""C_2/C_1` values would be minimum then, `log""0.01/0.1=log10^-1` (minimum).
Thus option (a) is CORRECT.


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