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Question
For the cell \[\ce{Mg_{(s)}|Mg^{2+}_{( aq)}||Ag^+_{( aq)}|Ag_{(s)}}\], calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell.
Given: \[\ce{E^0_{{Mg^{2+}|Mg}}}\] = −2.37 V and \[\ce{E^0_{{Ag^{+}|Ag}}}\] = 0.80 V
Solution
oxidation at anode
\[\ce{Mg -> Mg^{2+} + 2e^-}\]..........(1) `("E"_"ox"^0)` = 2.37 V
Reduction at cathode
\[\ce{Ag^+ + e^- -> Ag}\] ......(2) `("E"_"red"^0)` = 0.80 V
∴ `"E"_"cell"^0 = ("E"_"ox"^0)_"anode" + ("E"_"red"^0)_"cathode"`
= 2.37 + 0.80
= 3.17 V
Overall reaction
Equation (1) + 2 × equation (2)
\[\ce{Mg + 2Ag^{2+} -> Mg^{2+} + 2Ag}\]
∆G° = –nFE°
= –2 × 96500 × 3.17
= –611810 J
∆G° = –6.12 × 105 J
W = 6.12 × 105 J
∆G° = –2.303 RT log Kc
log Kc = `(6.12 xx 10^5)/(2.303 xx 8.314 xx 298)`
Kc = Antilog of (107.2)
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