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.

For more information about cell reaction, refer the link:

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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

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.