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.

A cell, `Ag|Ag^(+)||Cu^(2+)|Cu`, initially contains 1 M `Ag^(+)` and `

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» 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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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.

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