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Let us calculate the emf of the following cell at 25^@C using Nernst equation. Cu(s)|Cu^(2+)(0.25 aq. M)||Fe^(3+)(0.05 aq M)|Fe^(2+)(0.1 aq M)|pt(s) |
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Answer» Solution :Given : `(E^@)_(Fe^(3+)|Fe^(2+)) = 0.77 V and (E^@)_(Cu^(2+)|cu) = 0.34V` HALF reaction are `Cu(s) to Cu^(2+) (aq) + 2e^(-) "" ………(1)` `2Fe^(3+) (aq) + 2e^(-) to 2Fe^(2+)(aq)"" ……..(2)` the overall reaction is `Cu(s) + 2Fe^(3+)((aq)) to Cu^(2+)(aq) + 2Fe^(2+)(aq)`, and n = 2 APPLY NERNST equation at `25^@C` `E_("cell") = E_("cell")^(@) - (0.0591)/(2) log ([Cu^2][Fe^(2+)]^2)/([Fe^(3+)]^(2)) "" [ :. Cu(S) = 1]` `E_("cell")^(@) = (E_("OX")^(@))_(Cu|Cu^(2+)) + (E_("red")^(@))_(Fe^(3+)|Fe^(2+))` Given standared reduction potential of `Cu^(2+)|Cu` is 0.34 V `:. (E_("ox")^@)_(Cu|Cu^(2+)) = -0.34 V` `(E_("cell")^(@))_(Fe^(3+)|Fe^(2+)) = 0.77 V` `:. E_("cell")^(@) = -0.34 + 0.77 = 0.43 V` `:. E_("cell") = 0.43- (0.0591)/2 xx log ((0.25)(0.1)^2)/((0.005)^2)` `= 0.43 - (0.0591)/2 xx 2` `= 0.43 - 0.0591` `= 0.3709V` `= log((0.25)(0.1)^2)/((0.005)^2)` `= log(25 xx 10^(-2)xx 1 xx 10^(-2))/(25 xx 10^(-6))` `= log 10^2` `= 2 log_(10)10 = 2`. |
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