Question

1. What happens to the working electrode when the electrolyte concentration is changed?

2. Explain absorption emission through the Jablonski Diagram.

Answer 1

Effect of electrolyte concentration on working electrode is explained by the Nernst equation accodring to which the change in concentration affects the electrode potential in the following ways :

Ecell = Eocell - (RT/nF)/ln Q

This equation measures the Ecell under non standard conditions.

Let us consider and example of the galvanic cell set up at 298 K

Al(s) / Al3+(aq) //Ni2+(aq)/Ni(s)

Let us illustarte the effect of dilution on its Eocell

Eocell = Eocathode - Eoanode

= -0.25- (-1.66)

= 1.41 V

Under standard conditions, [Al3+] and [Ni2+] are 1 mol/dm^3

Case 1 Suppose when water is added to Ni2+/Ni half cell such that [Ni2+] is decreased 10 times. Then the Ecell is given as

Ecell = Eocell - (0.0592/n) logQ

Ecell = Eocell - (0.0592/n) log[Al3+]^2/[Ni2+]^3

= 1.41 - (0.0592/6) log 1/(1/10)^3

= 1.41 - 0.0296

= 1.38 V

Here Q >1 , Ecell decreases

Case 2 Suppose when water is added to Al3+/Al half cell such that [Al3+] is decreased 10 times. Then the Ecell is given as

Ecell = Eocell - (0.0592/n) logQ

Ecell = Eocell - (0.0592/n) log[Al3+]^2/[Ni2+]^3

= 1.41 - (0.0592/6) log (1/10)^2/(1)

= 1.41 - (-0.0197)

= + 1.43 V

Here Q < 1 , Ecell increases

Case 3 Suppose when water is added to both half cell such that [Al3+] and [Ni2+]is decreased 10 times. Then the Ecell is given as

Ecell = Eocell - (0.0592/n) logQ

Ecell = Eocell - (0.0592/n) log[Al3+]^2/[Ni2+]^3

= 1.41 - (0.0592/6) log (1/10)^2/(1/10)^3

= 1.41 - 0.00987

= +1.40 V

Q > 1, Ecell decreases a little

Therefore we can see how concentration of electrolyte afftects the electrode in terms of its potential ie

Any change inthe cell that increases Q decreases Ecell and vice versa.

Also adding reactant and removing product inceases Ecell and Adding product and removing reactant deceases Ecell.