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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)`

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Verified by Experts

The correct Answer is:
`E^(2)_("Cell")=E_("cathode")^(@)-E^(2)_("anode")=E_(Ag^(+)//Ag)^(@)-E_(Cu^(2+)//Cu)^(@)=+0.80-(+0.34)=0.46V`
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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]

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