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Determine the EMF of a decimolar Daniell...

Determine the EMF of a decimolar Daniell cell at `25^(@)C`. Given : `E_(Zn^(2+)|Zn)^(@)=-0.76V and E_(Cu^(2+)|Cu)^(@)=+0.34V`.

Text Solution

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The cell notation of a decimolar Daniell cell is :
`Zn|Zn^(2+)(0.1M)||Cu^(2+)(0.1M)|Cu`
`{:("Anode reaction: "Zn(s)" "rarr" "Zn^(2+)(aq)+2e),("Cathode reaction: "Cu^(2+)(aq)+2e" "rarr" "Cu(s)),(bar(" Cell reaction: "Zn(s)+Cu^(2+)(aq)" "rarr" "Zn^(2+)(aq)+Cu(s))):}`
2 mol of electrons are involved in the cell reaction, n = 2. EMF of the cell at `25^(@)C " "E_("cell")=E_("cell")^(@)-(0.059)/(2)log.([Zn^(2+)][Cu])/([Zn][Cu^(2+)])`
`=(E_("Cu^(2+)|Cu)^(@)-E_(Zn^(2+)|Zn)^(@))-(0.059)/(2)log.(0.1xx1)/(1xx0.1)`
`[because [Cu]=[Zn]=1]`
`=0.34-(-0.79)-(0.059)/(2)log1=1.10V`
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