Question

calculate delta G

Calculate AGA° for each of the following reactions. Use the standard reduction potentials in these Reference Tables. Enter your answers to three significant figures. 6 Fe2 (aq) 2 Cr3+(aq) (a) Cr2072-(aq) 14 H1(aq) 6 Fe3(aq) 7 H20(I) kJ (b) 3 Pb2+(aq) + 2 Cr3+(aq) 7 H20(I) 3 Pb(s) Cr2072(aq) + 14 H1 (aq) kJ In which directions are these reactions spontaneous? (Select all that apply.) OReaction (a) - forward Reaction (a) - reverse Reaction (a) - neither direction Reaction (b) - forward Reaction (b) - reverse Reaction (b) neither direction

Answer 1

ΔG⁰ of a reaction is called the change in free energy for that reaction. Now it is calculated from the formula-

ΔG⁰ = -nFE⁰

where n = number of moles of electrons involved in the reaction

F = faradeys constant = 96485 C

E⁰ = Standard reduction potential of cell

1- For the given reaction

Cr₂O₇^2- + 14 H⁺ + 6Fe⁺² -------------> 2Cr⁺³ + 6Fe⁺³ + 7H₂O

Now we know in an electrochemical cell, two half reactions occurs.

• Oxidation - i.e loss of electron
• Reduction - i.e gain of electron

The E⁰ of the cell =  E⁰ (reduction half cell) - E⁰ (oxidation half cell)

Now for the above reaction, we can see the oxidation state of Cr in the left = +6 and in the right = +3. That means gain of electron take place here. So Cr has undergone Reduction

And oxidation state of Fe in the left = +2 and in the right = +3. That means loss of electron take place here. So Fe has undergone Oxidation

Hence

Oxidation half cell : Fe⁺² -------------> Fe⁺³ + e E⁰_red =  -0.04 V

Reduction half cell : Cr⁺⁶ + 3e-------------> Cr⁺³ E⁰_red = 1.33 V

So balancing these reactions

Oxidation half cell : Fe⁺² -------------> Fe⁺³ + e]*3 E⁰_red =  -0.04 V * 3

Reduction half cell : Cr⁺⁶ + 3e-------------> Cr⁺³ E⁰_red = 1.33 V

Or

Oxidation half cell : 3Fe⁺² -------------> 3Fe⁺³ + 3e E⁰_red =  -0.12 V

Reduction half cell : Cr⁺⁶ + 3e -------------> Cr⁺³    E⁰_red = 1.33 V

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Overall reaction :  3Fe⁺² + Cr⁺⁶ -------------> 3Fe⁺³ + Cr⁺³ E⁰cell = E⁰_red (1.33 V) -  E⁰_red (-0.12 V)

= 1.33 V + 0.12 V

= 1.45 V

So putting the values

ΔG⁰ = -nFE⁰

= -3 * 96485 C * 1.45 V

= -419709 J

= -419.709 kJ

Now since the calculated ΔG⁰ has a negative value, that means the reaction (a) is spontaneous in forward direction.

Similarly for reaction- b

3Pb²⁺ + 2Cr⁺³ + 7 H2O -------------> 3Pb + Cr₂O₇^2- + 14 H⁺

Oxidation half cell : Cr⁺³ -------------> Cr⁺⁶ + 3e E⁰_red =  -1.33 V

Reduction half cell : Pb⁺² + 2e-------------> Pb E⁰_red = -0.13 V

So balancing these reactions

Oxidation half cell : Cr⁺³ -------------> Cr⁺⁶ + 3e]*2 E⁰_red =  -1.33 V * 2

Reduction half cell : Pb⁺² + 2e-------------> Pb] *3 E⁰_red = -0.13 V * 3

Or

Oxidation half cell :   2Cr⁺³ -------------> 2Cr⁺⁶ + 6e E⁰_red =  -2.66 V

Reduction half cell : 3Pb⁺² + 6e-------------> 3Pb    E⁰_red = -0.39 V

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Overall reaction :  2Cr⁺³ + 3Pb⁺² -------------> 2Cr⁺⁶ + 3Pb E⁰cell = E⁰_red (-0.39 V) -  E⁰_red ( -2.66 V)

= -0.39 V + 2.66 V

= 2.27 V

So putting the values

ΔG⁰ = -nFE⁰

= -6 * 96485 C * 2.27 V

= -13141259 J

= -13141.259 kJ

Now since the calculated ΔG⁰ has a negative value, that means the reaction (b) is spontaneous in forward direction.