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A voltaic cell is set up at 25^(@)C with...

A voltaic cell is set up at `25^(@)C` with the following half cells :
`Ag^(+) (0.001M)` | Ag and `Cu^(2+)(0.10 M)|Cu`
What would be the voltage of this cell ? `(E_("cell")^(@) = 0.46 V)`

Text Solution

Verified by Experts

`E_("cell")^(Theta)=0.46V,T=25+273=298K`
The cell reaction for the given cell may be represented as:
`Cu(s) to Cu^(2+)(aq) +2e^(-)` (Oxidation half reaction)
`2Ag^(+) (aq) + 2e^(-) to 2Ag(s)` (Reduction half-reaction)
Net-reaction. `Cu(s) + 2Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag (s)`
Thus, n=2
Applying the Nernst equation,
`E_("cell")=E_("cell")^(Theta)-(0.059)/(2)"log"([Cu^(2+)]^(2))/([Ag^(+)]^(2))`
`=0.46 -0.0295"log"([0.01])/([0.001]^(2))`
`=0.46-0.0295"log"10^(5)`
`=0.46-0.0295 xx 5=0.3125V`
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