Calculate emf and ΔG for the following cell at 298 K:
Mg(s) | Mg2+(0.01 M) || Ag+(0.001 M) | Ag(s)
Given: E°Mg2+/Mg = -2.37 V, E°Ag+/Ag = +0.80 V. [1 F = 96500 C mol-1, log 10 = 1]
Mg(s) | Mg2+(0.01 M) || Ag+(0.001 M) | Ag(s)
Given: E°Mg2+/Mg = -2.37 V, E°Ag+/Ag = +0.80 V. [1 F = 96500 C mol-1, log 10 = 1]
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Solution and Explanation
Answer:
Standard cell potential: E°cell = E°cathode - E°anode = 0.80 - (-2.37) = 3.17 V.
Cell reaction: Mg + 2 Ag+ → Mg2+ + 2 Ag, so n = 2.
Nernst equation at 298 K: Ecell = E°cell - (0.0591/n) log ([Mg2+] / [Ag+]2).
Reaction quotient = [Mg2+]/[Ag+]2 = 0.01 / (0.001)2 = 0.01 / 0.000001 = 104.
log(104) = 4.
Ecell = 3.17 - (0.0591/2)(4) = 3.17 - (0.02955)(4) = 3.17 - 0.1182 = 3.05 V (approximately 3.052 V).
ΔG = -nFEcell = -(2)(96500)(3.05) = -588,650 J = about -588.6 kJ (approximately -5.89 × 105 J/mol). The negative sign shows the cell reaction is spontaneous.
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