1.

For the Mg-Ag cell, how many times the difference between the EMF of the cell and its standard EMF will change if concentration of Mg^(2+) ions is changed from 0.1 M to 0.01 M and that of Ag^(+) ions is chagned from 0.5 M to 0.25 M?

Answer»


Solution :`Mg+2AG^(+)toMg^(2+)+2Ag,E_(CELL)=E_(cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)])/([Ag^(+)]^(2))`
or `E_(cell)^(@)-E_(cell)=(0.0591)/(n)"log'([Mg^(2+)])/([Ag^(+)]^(2))`
`therefore` In 1st CASE, `E_(cell)^(@)-E_(cell)=(0.0591)/(2)"log"(0.1)/((0.5)^(2))`
In 2nd case, `E_(cell)^(@)-E_(cell)=(0.0591)/(2)"log"(0.01)/((0.25)^(2))`
`=(0.0591)/(2)"log((0.1)^(2))/([(0.5)^(2)]^(2))`
`=(0.0591)/(2)xx2log(0.1)/((0.05)^(2))=2` TIMES


Discussion

No Comment Found

Related InterviewSolutions