Question

Determine the standard potential (at 25 ° C) of the cell: A (s) | A5 + (ac) || ​​B3 + (ac) | B (s), if [AS +] = 1.91 M and [B³ *] = 0.81 M. Use the following standard reduction potentials.

reduction reaction. E ° (V)

A5 + (ac) + 5 e- A (s) 0.039

B3 + (ac) + 3 e - B (s) 0.577

Answer 1

We know in an electrochemical cell both Oxidation (i.e loss of electron) and Reduction (i.e gain of electron) takes place.

Now for the given electrochemical cell A (s) | A5+ (aq) || ​​B3+ (aq) | B (s)

The Half cells are-

Oxidation : A (s) ------------> A5+ (aq) + 5e Eored = 0.039 V

Reduction : B3+ (aq) + 3e ------------> B (s) Eored = 0.577 V

Now to calculate the standard potential (Eocell) we need to make the number of electrons transfered equal for both the half cells. So for this we need to multilply 3 in the Oxidation half and 5 in Reduction half and add both i.e

Oxidation : A (s) ------------> A5+ (aq) + 5e Eored = 0.039 V ] * 3

Reduction : B3+ (aq) + 3e ------------> B (s) Eored = 0.577 V ] * 5

Or

Oxidation : 3A (s) ------------> 3A5+ (aq) + 15e Eored = 0.117‬ V

Reduction : 5B3+ (aq) + 15e ------------> 5B (s) Eored = 2.885 V

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Overall :  3A (s) + 5B3+ (aq) ------------> 3A5+ (aq) + 5B (s) Eocell = Eored (Red half) - Eored (Oxd half)

= 2.885 V - 0.117‬ V

= 2.768‬ V

Now the cell potential (Ecell) at 25 ° C is calculated as per the Nernst equation-

Ecell = Eocell - 0.0592/n * log Q

Where

n = number of electrons transfered = 15

Q = Reaction quotient

= [A5+ (aq)]3 / [B3+ (aq)]5  

= [1.91]3 / [0.81]5  

= 19.98

So putting the values,

Ecell = Eocell - 0.0592/n * log Q

=  2.768‬ V - 0.0592/15 * log (19.98)

= 2.768‬ V - (0.0592/15 * 1.3)

= 2.768‬ V - 0.005

= 2.763‬ V