## HCl as anode and cathode in cell diagram

Hannah_1G
Posts: 100
Joined: Wed Sep 18, 2019 12:22 am

### HCl as anode and cathode in cell diagram

How does this cell diagram work Pt(s)|H 2 (g, 1.0 bar)|HCl(aq, 0.075 mol/L )||HCl(aq, 1.0 mol/L)|H 2 (g, 1.0 bar)|Pt(s)
and how would I balance it to solve for its cell potential?

Aliya Jain 2B
Posts: 101
Joined: Wed Sep 11, 2019 12:16 am

### Re: HCl as anode and cathode in cell diagram

I think this means that is a concentration cell. In this case you would use the nernst equation, but e standard is 0.

Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

### Re: HCl as anode and cathode in cell diagram

You would take the half reactions:
2H+ (aq,1.0M) + 2e- --> H2 (g,1atm) // E cathode = 0.00V
H2 (g,1atm) --> 2H+ (aq,0.075M) + 2e- // E anode = 0.00V

2H+ (aq, 1.0M) + H2 (g,1atm) --> 2H+ (aq,0.075M) + H2 (g,1atm)
and get Ecell = 0.00V

Then use the nernst equation to get:
$E=E^\circ -(\frac{0.025693V}{n})ln(\frac{[H^+,0.075M]^2P_H_2}{[H^+,1.0M]^2P_H_2})$
Where the numerator in the ln function represents the products and the denominator represents the reactants of the overall balanced rxn

And then solve for E

Hannah_1G
Posts: 100
Joined: Wed Sep 18, 2019 12:22 am

### Re: HCl as anode and cathode in cell diagram

Julie Park 1G wrote:You would take the half reactions:
2H+ (aq,1.0M) + 2e- --> H2 (g,1atm) // E cathode = 0.00V
H2 (g,1atm) --> 2H+ (aq,0.075M) + 2e- // E anode = 0.00V

2H+ (aq, 1.0M) + H2 (g,1atm) --> 2H+ (aq,0.075M) + H2 (g,1atm)
and get Ecell = 0.00V

Then use the nernst equation to get:
$E=E^\circ -(\frac{0.025693V}{n})ln(\frac{[H^+,0.075M]^2P_H_2}{[H^+,1.0M]^2P_H_2})$
Where the numerator in the ln function represents the products and the denominator represents the reactants of the overall balanced rxn

And then solve for E

Thank you! So when the same compounds are used as both anode and cathode, the standard cell potential is always 0?