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For the cell Mg(s) |Mg^(2+)(aq)||Ag^(+)(aq)|Ag (s), calculate the equilibrium constant at 25^@C and maximum work that can be obtained during operation of cell. Given : E_(Mg^(2+)|Mg)^@ = + 2.37 V and E_(Ag^(2+)|Ag)^@ = 0.80V |
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Answer» Solution :`{:("Oxidation at anode", Mg to Mg^(2+)+2e^(-), (E_("ox")^(@)) = 2.37V),("Reduction at cathode ", Ag^+ + e^(-) to Ag, (E_("red")^@) = 0.80 V):}` `:. E_("cell")^(@) = (E_("ox")^(@))_("anode") + (E_("red")^(@))_("cathode") = 2.37 + 0.80 = 3.17V` Overall reaction, `Mg + 2Ag^(+) to Mg^(2+) +2Ag` `DeltaG^@ = -NFE^@` `= -2 XX 96400 xx 3.17` `= -6.118 xx 10^5J` We know that `W_("MAX") = DeltaG^@` `:. W_("max") = +6.118 xx 10^5 J` Relationship between `DeltaG^@` and `K_(eq)` is `Delta G = -2.303 RT log K_(eq)` `DeltaG = -2.303 xx 8.314 xx 298 log K_(eq) "" [ :. 25^@C = 298 K]` `log K_(eq) = (6.118 xx 10^5)/(2.303 xx 8.314 xx 298) IMPLIES (6.118 xx 10^5)/(5705.84)` `log K_(eq) = 107.223` `K_(eq) = "Antilog" (107.223)` . |
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