1.

If the half cell reactions are given as (i) Fe_((aq))^(2+)+2etoFe_((s)),E^(o)=0.44V (ii) 2H_((aq))^(+)+(1)/(2)O_(2(g))+2etoH_(2)O_((l)),E^(o)=+1.23V The E^(o) for the reaction Fe_((s))+2H^(+)+(1)/(2)O_(2(g))toFe_((aq))^(2+)+H_(2)O_((l))

Answer»

`+1.67V`
`-1.67V`
`+0.79V`
`-0.79V`

Solution :Subtracting EQUATION (i) from (II) we will get equation (iii)
`Fe_((s))+2H^(+)+(1)/(2)O_(2)toFe_((aq))^(2+)+H_(2)O_((l))`. . .(iii)
So, `E^(o)=E_(2)^(o)--E_(1)^(0)`
`=+1.23-(-0.44)=+1.6V`


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