1.

A cell, Ag|Ag^(+)||Cu^(2+)|Cu, initially contains 1 M Ag^(+) and 1M" "Cu^(2+) ions. Calculate the change in cell potential after the passage of 9.65 A of a current of 1 h.

Answer»

SOLUTION :Quantity of electricity passed passed=`9.65xx60xx60C=34740C`
`thereforeAg^(+)` ions DEPOSITED`=34740//96500" mole"=0.36` mole`(Ag^(+)+e^(-)toAg)`
`Cu^(2+)` ions deposited`=34740//(2xx96500)=0.18` mole`(Cu^(2+)+2e^(-)toCu)`
`therefore[Ag^(+)]"left"=1-0.36=0.64M`
`[Cu^(2+)]" left"=1-0.18=0.82M`
Cell reaction is: `Cu+2AG^(+)toCu^(2+)+2Ag`
`DeltaE=E_(cell)^(@)=(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))=(0.0591)/(2)"log"(0.82)/((0.64)^(2))=8.89xx10^(-3)V`


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