Question

Consider the following: Au" + 30 Ag+ + le à Au à As e - 142 V E = 0.80 V Which specie is the best oxidizing agent? Which specie is being oxidized?

Answer 1

IN an eletcrochemical cell, there are 2 half cells. In one oxidation (i.e loss of electron) and in another Reduction (i.e gain of electron) takes place

Now the cell potential is generally written in Reduction potential term.

And the cell, which has the higher Reduction potential value is the actual Reduction half cell and the cell which has the lower Reduction potential value is the Oxidation half cell

So between the given cells,

Au+3 + 3e ------> Au E⁰ = 1.42 V

Ag+ + 1e ---------> Ag E⁰ = 0.80 V

Here Au+3 has the higher Reduction potential value. So

• Au will be reduced i.e Au+3 + 3e ------> Au
• Ag will be oxydised. i.e Ag ------> Ag+ + 1e

Now the species which will get itself reduced is the Oxydising agent and the species which will get itself oxidized is the Reducing agent

So here

• Oxidizing agent = Au
• Reducing agent = Ag

9-

Now standard cell potential is

E⁰cell = E⁰reduction - E⁰oxidation

So for

• Au+3 + 3e ------> Au E⁰ = 1.42 V
• Ag+ + 1e ---------> Ag E⁰ = 0.80 V

Balancing these

• Au+3 + 3e ------> Au E⁰ = 1.42 V
• Ag ---------> Ag+ + 1e ]*3 E⁰ = 0.80 V * 3

Or

• Au+3 + 3e ------> Au E⁰ = 1.42 V
• 3Ag ---------> 3Ag+ + 3e E⁰ = 2.4 V

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• Au+3 + 3Ag -------------> Au + 3Ag+   E⁰ = 1.42 V - 2.4 V = -0.98‬ V

10-

Now at +ve electcrode, all the -ve ions will be attracted. That means this electrode can accept electrons from those negative ions. So it act as an oxydizing Agent. So the species present here is Au

11-

Now the gibbs free energy (ΔG) = -nFE⁰

Where n = number of electrons involved = here n = 3

F = faradeys constant = 46485 C

So

ΔG = -nFE⁰

= -3 * 46485 C * (-0.98‬ V)

= 136666 J