Question

The cell potential for the cell Zn(s) + 2H+(? M) LaTeX: \longrightarrow⟶ Zn2+(1.3 M) + H2(g) (8 atm) is observed to be 0.68 V. What is the pH in the H+/H2 half-cell? Reduction potential for H2(g)/H+(aq) is 0.00 V, for Zn(s)/Zn2+(aq) is -0.76 V. Enter number to 2 decimal places.

Answer 1

pH in the
`H_(2)/H^(+)` half cell is 0.84.

Step-by-step explanation:

Oxidation:
`Zn(s)-2e^(-)\rightarrow Zn^(2+)(aq.)`

Reduction:
`2H^(+)(aq.)+2e^(-)\rightarrow H_(2)(g)`

-------------------------------------------------------

Overall:
`Zn(s)+2H^(+)(aq.)\rightarrow Zn^(2+)(aq.)+H_(2)(g)`

`E_(cell)^(0)=E_{H^(+)\mid H_(2)}^(0)-E_(Zn^(2+)\mid Zn)^(0)` = (0.00 V) + (0.76 V) = 0.76 V

According to Nernst equation for this cell reaction at room temperature (298 K):

`E_(cell)=E_(cell)^(0)-(0.0592)/(n)log\frac{[Zn^(2+)].P_{H_(2)}}{[H^(+)]^(2)}`

where,
`E_(cell)` is cell potential, n is number of electron exchanged,
`P_{H_(2)}` is pressure of
`H_(2)` in atm and species under third bracket represent molarity of the respective species.

So,
`0.68V=0.76V-(0.0592)/(2)log((1.3M)* (8atm))/([H^(+)]^(2))V`

`\Rightarrow`
`[H^(+)]=0.1436M`

pH =
`-log[H^(+)]` = -log(0.1436) = 0.84