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

Answer 1

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 K_c

log K_c = `(6.12 xx 10^5)/(2.303 xx 8.314 xx 298)`

K_c = Antilog of (107.2)