How Does Electrolysis Affect pH in a HBr Solution?

[harrietstowe]

Homework Statement

A current of 10.7 Amps is applied to a 1.30 L solution of 0.550M HBr converting some of the H+ to H2(g), which bubbles out of solution. What is the PH of the solution after 77 minutes.

Homework Equations

The Attempt at a Solution

(10.7 C/s)(1 mol e/96485 C)(2 mol H+/2 mol e)(60s/min)(77 min) = .512 mol H+

.512 mol H+/1.30L = .394M H+
-log(.394M) = .405 = PH

This was incorrect

 

[pmsrw3]

You started with a certain amount of H+ in the solution. The electrolysis removed some. Now you need to calculate what's left.

 

[harrietstowe]

I see will try, thank you

 

[harrietstowe]

It worked, thank you

 

[DeeNos]

because the concentration of H+ has changed due to the conversion to H2 gas. In order to accurately calculate the pH, we need to consider the new concentration of H+ ions in the solution. This can be done by using the Nernst equation, which takes into account the effect of the gaseous H2 on the concentration of H+ ions in the solution. The Nernst equation is as follows:

Ecell = E°cell - (RT/nF)lnQ

Where:
Ecell = cell potential
E°cell = standard cell potential
R = gas constant (8.314 J/mol*K)
T = temperature (in Kelvin)
n = number of electrons transferred in the reaction
F = Faraday constant (96,485 C/mol)
Q = reaction quotient (concentration of products over concentration of reactants)

In this case, the reaction is:

2HBr(aq) → H2(g) + 2Br-(aq)

The reaction quotient, Q, can be expressed as:

Q =

/[HBr]^2

Since we know that the concentration of H2 is 0.550M (from the given information) and the concentration of HBr is 0.550M before the reaction and 0.156M after the reaction (calculated from the given information), we can substitute these values into the equation:

Q = (0.550M)/(0.156M)^2 = 2.817

We also know that the standard cell potential for this reaction is 0V, so the Nernst equation becomes:

Ecell = 0 - (8.314 J/mol*K)(298K)/(2 mol)(96,485 C/mol)ln(2.817) = -0.029V

From the Nernst equation, we can see that the cell potential is now negative, indicating that the reaction is not spontaneous and the concentration of H+ ions has decreased. This means that the pH of the solution has increased. To calculate the new pH, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:
pH = -log([H+])
pKa = -log(Ka) (Ka is the acid dissociation constant)
[A-] = concentration of conjugate base (in this case, Br-)
[HA] = concentration of acid (in this case,