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A voltic cell is set up at 25^@C with th...

A voltic cell is set up at `25^@C` with the half cells `Ag^(+) (0.001M)` Ag and `Cu^(2+)(0.10M)` . What should be its cell potential.[ E c u = + 0.34 V , E ∘ A g = + 0.8 V ]

Text Solution

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`Cu+2Ag^(+) to Cu^+ +2Ag`
Half cell reactions:
Cathade (reduction )
`2Ag^+ (0.001M)+2e^(-) to 2Ag(s) `
Anode (oxidation):
`Cu_((s))to Cu^(2+)(0.10M)+2e^-`
`therefore n=2`
`E_(cell)^@=0.46V`
`E_(cell)=E_(cell)^@-0.059/2log""([Cu^(2+)])/([Ag^+]^2)`
`=0.46-0.059/2log""((0.1))/(0.001)^2`
`=0.46-0.059/2log 10^5`
`=0.46 -0.059/2 times (5)`
`=0.46-0.1475`
=0.3125V
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