1.

Calculate the reduction potential of a half-cell consisting of platinum electrode immersed in 2.0 M Fe^(2+) and 0.02 M Fe^(3+). Given E_(Fe^(3+)//Fe^(2+))^(@)=0.771V

Answer»

0.653 V
0.889 V
0.683 V
2.771 V

Solution :`FE^(3+)+E^(-)toFe^(2+)`
`E_(RED)=E_(Fe^(2+)//Fe^(2+))^(@)-(0.0591)/(2)"log"([Fe^(2+)])/([Fe^(3+)])`
`=0.771-0.0591"log"(2.0)/(0.02)`
`=0.771-0.591xxlog100`
`=0.771-0.0591xx2=0.653V`.


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