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))`A. `+1.67V`B. `-1.67V`C. `+0.79V`D. `-0.79V`

If the half cell reactions are given as (i) `Fe_((aq))^(2+)+2etoFe_((s

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))`A. `+1.67V`B. `-1.67V`C. `+0.79V`D. `-0.79V`
Answer» Correct Answer - A 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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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))`A. `+1.67V`B. `-1.67V`C. `+0.79V`D. `-0.79V`

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