Calculate E o , E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0.025 M and [Sn2+] = 0.040 M E o = V E = V ΔG = kJ (b) 3Zn(s) + 2Cr3+(aq) ⇌ 3Zn2+(aq) + 2Cr(s) where [Cr3+] = 0.040 M and [Zn2+] = 0.0085 M E o = V E = V ΔG = kJ

(a) = 2.23 V, E = 2.2 V, G = 424.6 kJ/mol

(b) = 0.336 V, E = 0.332 V, G = 64.076 kJ/mol

(a) Oxidation potential of Mg = 2.37

Oxidation potential of = -0.14

Mg + = 2.37 + (-0.14)

= 2.23 V

According to nerst equation:

E = 2.2 V

G= nFE

G = 2 96500 2.2

G = 424600 J/mol

G = 424.6 kJ/mol

(b) Reduction potential of Zn = 0.76

Reduction potential of Cr = 0.424

= reduction - oxidation

= 0.76 - 0.424

= 0.336 V

According to nerst equation:

E = 0.332 V

G= nFE

G = 2 96500 0.332

G = 64076 J/mol

G = 64.076 kJ/mol.

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E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

Explanation:

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

a:

by the combination of an unlimited number of elements