1.

Calculate the reduction potential of a half-cell containing of platinum electrode immersed in 2.)Mfe^(2+) and 0.2MFe^(3+), Given:- E_(Fe^(3+)//Fe^(2+))^(o)=0.771V Fe^(3+) +e^(-)toFe^(2+)

Answer»

0.653V
0.889V
0.683V
2.771V

Solution :`E_(CELL)=E_(cell)^(0)-(RT)/(NF)xx2.303"log"([Fe^(2+)])/([Fe^(3+)])`
`E_(cell)=0.771-(8.314xx2.303xx298)/(96500)"log"(2)/(0.02)`
`=0.771-(0.0591xx2)=0.771-0.1182`
`E_(cell)=0.653V`.


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