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A cell is set up between copper and silv...

A cell is set up between copper and silver electrodes as follows:
`Cu(s)I Cu^(2+)(aq)II Ag^+(aq)Iag(S)`
If the two half cells work under standard conditions, calculate the EMF of the cell
`(Given E^(@)_(Cu^(2+)//Cu) =+0.34 V,E^(@)_(Ag^(+)//Ag)" " =+0.80 V)`

Text Solution

Verified by Experts

From the `E^(@)` values, it is clear that copper acts as the anode and silver as the cathode.
`:.E_(cell)^(@)=E_(cathode)-E_(anode)^(@)=0.80-(0.34)=+0.46 " volt "`
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Explore conceptually related problems

Consider a cell composed of two cells: (i). Cu(s)Cu^(2+)(aq) and (ii). Ag(s)|Ag^(+)(aq) (b). The cell potential when [Cu^(2+)]=2M and |Ag^(+)|=0.05M [Given: C_(Cu^(2+)//Cu)^(@)=+0.344V,E_(Ag^(+)//Ag)^(@)=+0.80V]

Calculate the emf of the following cell: Cu(s)|Cu^(2+) (aq)||Ag^(+) (aq)| Ag(s) Given that, E_(Cu^(2+)//Cu)^(@)=0.34 V, E_(Ag//Ag^(+))^(@)=-0.80 V

Knowledge Check

  • In the reaction , Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2Ag(s) , the reduction half cell reaction is

    A
    `Cu+2e^(-)rarrCu^(2+)`
    B
    `Cu-2e^(-)rarrCu^(2+)`
    C
    `Ag^(+)+e^(-)rarrAg`
    D
    `Ag^(-)e^(-)rarrAg^(+)`

  • Calculate the emf of the following cell: Cu(s)|Cu^(2+) (aq)||Ag^(+) (aq)| Ag(s) Given that, E_(Cu^(2+)//Cu)^(@)=0.34 V, E_(Ag//Ag^(+))^(@)=-0.80 V

    A
    0.046 V
    B
    0.46 V
    C
    0.57 V
    D
    `-0.46 V`
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