1.

If Zn^(2+)//Zn electrode is diluted 100 times, then the change in emf is

Answer»

increase of 59 mV
decrease of 59 mV
increase of 29.5 mV
decrease of 29.5 mV

Solution :`E_(cell)^(@)=E_(cell)^(@)-(0.0591)/(2)"LOG"(1)/([ZN^(2+)])=E_(cell)^(@)+0.02955logC`
`E_(cell)'=E_(cell)^(@)+0.02955"log"(C)/(100)=E_(cell)^(@)+0.02955(logC-2)`
`thereforeE_(cell)` will decrease by 0.02955`xx2V=0.059V=59mV`.


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