1.

giventhe standardhalf-cellpotentials(E^@)of thefollowing as :Zn=Zn^(2+) +2e^(- ), E^@= + 0.76 V Fe = Fe^(2+) + we^(-) , E^@= 0.41V thenthe standarde,m,fof thecellwith thereaction

Answer»

`-0.35 V`
`+0.35 V`
`+1.17V`
`-1.17 V`

Solution :`FE^(2+) +Znto Fe+Zn(2+)`
`THEREFORE ` StandardEMFof thecell
= Oxidationpotentialof anode+ Reductionpotentialofcathode
`-(0.76) + ( -0.41 ) = +0.35V `


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