1.

If `Zn^(2+)//Zn` electrode is diluted 100 times, then the charge in reduction potential isA. increase of 59mVB. decrease of 59mVC. increase of 25.5mVD. decrease of 2.95V

Answer» Correct Answer - B
`E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.059)/(2)"log"(1)/(|Zn^(2+)|)`
`E_(Zn^(2+)//Zn)=E_(Zn^(3+)//Zn)+0.0295"log"C`
When solution is diluted 100 times.
`E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)+0.0295"log"(C)/(100)`
`=E_(Zn^(2+)//Zn)^(@)+0.0295"log" C-0.0295xx2`
`=E_(Zn^(2+)//Zn)-0.059`
`E_(Zn^(2+)//Zn)-E_(Zn^(3+)//Zn)=-0.059V=-59mV`


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