1.

The standard reduction potential E^(@) for the half reactions are as Zn=Zn^(2+)+2e^(-),E^(@)=+0.76V Fe=Fe^(2+)+2e^(-),E^(@)=+0.41V The EMF for cell reaction Fe^(2+)+ZntoZn^(2+)+Fe is

Answer»

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

SOLUTION :In this reaction.
EMF=`E_("cathode")-E_("anode")=-0.41-(-0.76)" EMF"=+0.35V`.


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