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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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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^(+)`

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