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Calculate the standard free-energy change for the reaction at 25 ∘c. refer to the list of standard reduction potentials. 2au3 (aq) 3cr(s)↽−−⇀2au(s) 3cr2 (aq)

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tushargupta0691

The standard free-energy change for the reaction at 25°C is 92640 KJ.

The change in free energy that happens when a compound is created from its constituent parts in their most thermodynamically stable states under standard-state circumstances is known as the standard-state free energy of creation.

There are 6 electrons transferred. The anode half-reaction produces 6 electrons while the cathode half-reaction uses 6 electrons.

For these types of reactions always write out the half-reactions and the appropriate oxidation numbers.

6 electrons are transferred as Au³+ has an oxidation number of 3+ and Au has an oxidation number of 0. 3 electrons are gained for each Au³+ and there are 2 Au³+ in the equation

                         [tex]2Au^3 (aq) +3cr(s)[/tex] ↽−−⇀ [tex]2Au(s) +3Cr^2[/tex]

Same for Cr to Cr2+ because there are 3 moles of it.

So, n = 6.

     F = 96500 J/(V*mol)

     where  F is faraday's constant.

                 n = moles of electrons are transferred in this reaction.

Standard emf of the cell, E = cathode - anode.

                                             = 1.49 - 1.33

                                             = 0.16

Using the equation,

                                   ΔG = -nFE

                                         = 6 × 96500 × 0.16

                                         = 92640 KJ

Therefore, the standard free-energy change for the reaction at 25°C is 92640 KJ.

Learn more about standard free energy here:

https://brainly.com/question/10012881

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