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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.[Ecu=+0.34V,E∘Ag=+0.8V] |
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Answer» Solution :`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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