E⁰(cell) = 2.24V
E(cell) = 2.246V
∆G = -433 KJ/mol
<u>Explanation:</u>
Mg(s) + Sn²⁺(aq) ⇌ Mg²⁺(aq) + Sn(s)
[Mg2+] = 0.025 M
[Sn2+] = 0.040 M
First we need the standard reduction potentials:
. . . . . . . . . . . . . . . . . E°(V)
Mg²⁺ + 2 e⁻ ⇌ Mg(s). . .−2.372
Sn²⁺ + 2 e⁻ ⇌ Sn(s) . . . −0.13
Take the more negative (or less positive in other cases) one, and write it as an oxidation:
Mg(s) ⇌ Mg²⁺ + 2 e⁻. . .+2.372 V
Combine them,
Mg(s) + Sn²⁺ ⇌ Mg²⁺ + Sn(s)
E°(cell) = +2.372 – 0.13 V = 2.24 V
To get the cell potential under the conditions given, use the Nernst Equation:
E(cell) = E°(cell) – [(0.059)/n]•logQ = 2.24 V – 0.0295 V • log [Mg²⁺]/[Sn²⁺]
Note that the solids don't appear in Q, only the concs. of the dissolved ions.
E(cell) = 2.24 V – 0.0295 V X log (0.025)/(0.040)
= 2.24 + 0.006 V ≈ 2.246 V
The concentration ratio in Q (Sn²⁺ and Mg²⁺) is too close to 1 to shift E(cell) significantly from E°(cell) given the precision I have for the Sn reduction potential.
∆G = –nFE(cell) = –2(96.485 kJ/mol•V)(2.246 V) = –433 kJ/mol
E⁰(cell) = 2.24V
E(cell) = 2.246V
∆G = -433 KJ/mol
