[Abhinab]

Calculate the concentration of H⁺ .

Hydrogen electrode at 1 atm is connected with Zn electrode in which the emf(E) of the cell is found to be 0.54 V at 25⁰C. Suppose that [Zn⁺⁺]=1.0M. Calculate the molar concentration of H⁺.

[sjb]

Calculate the concentration of H⁺ .

Hydrogen electrode at 1 atm is connected with Zn electrode in which the emf(E) of the cell is found to be 0.54 V at 25°C. Suppose that [Zn⁺⁺]=1.0M. Calculate the molar concentration of H⁺.

Any equations you know that may help?

[Abhinab]

Using standard electrode potential of zn/zn++ as -0.76V, we can find the E⁰_cell and we can use Nernst Equation to find the molar concentration as;

E_cell=E⁰_cell-(0.0591/n)log₁₀[Q]
where Q={[Zn⁺⁺]/[H⁺]}
 

[mjc123]

What is n? and what is the correct expression for Q? Try writing out the equation for the cell reaction.

[Abhinab]

at anode;
Zn Zn⁺⁺ + 2e^-    E⁰=0.76V
at cathode;
2H⁺ + 2e^-  H₂     E⁰=0V
Net cell reaction:
Zn + 2H⁺  Zn⁺⁺ + H₂     E⁰_cell=0.76V
Therefore,
E_cell=E⁰_cell-(0.0591/n)log(Q)
or, 0.54=0.76-(0.0591/2)log{([H₂]x[zn⁺⁺])/([Zn]x[H⁺]²)}
where, [Zn]=[H₂]=[Zn⁺⁺]=1
Therefore, [H⁺]=1.89x10^-4 M

[Borek]

Zn Zn⁺⁺ + 2e^-    E⁰=0.76V

-0.76V

Therefore, [H⁺]=1.89x10^-4 M

Wasn't that hard.

Note, that most of the work you did by treating whole system was unnecessary. You could treat Zn electrode just as a reference. As [Zn²⁺] = 1M electrode in the standard state and its potential is -0.76V. That means potential of the hydrogen electrode is just -0.22V (vs standard hydrogen electrode), and what you have is just a simple concentration cell.