Answer:
pH in the [tex]H_{2}/H^{+}[/tex] half cell is 0.84.
Explanation:
Oxidation: [tex]Zn(s)-2e^{-}\rightarrow Zn^{2+}(aq.)[/tex]
Reduction: [tex]2H^{+}(aq.)+2e^{-}\rightarrow H_{2}(g)[/tex]
-------------------------------------------------------
Overall: [tex]Zn(s)+2H^{+}(aq.)\rightarrow Zn^{2+}(aq.)+H_{2}(g)[/tex]
[tex]E_{cell}^{0}=E_{H^{+}\mid H_{2}}^{0}-E_{Zn^{2+}\mid Zn}^{0}[/tex] = (0.00 V) + (0.76 V) = 0.76 V
According to Nernst equation for this cell reaction at room temperature (298 K):
[tex]E_{cell}=E_{cell}^{0}-\frac{0.0592}{n}log\frac{[Zn^{2+}].P_{H_{2}}}{[H^{+}]^{2}}[/tex]
where, [tex]E_{cell}[/tex] is cell potential, n is number of electron exchanged, [tex]P_{H_{2}}[/tex] is pressure of [tex]H_{2}[/tex] in atm and species under third bracket represent molarity of the respective species.
So, [tex]0.68V=0.76V-\frac{0.0592}{2}log\frac{(1.3M)\times (8atm)}{[H^{+}]^{2}}V[/tex]
[tex]\Rightarrow[/tex] [tex][H^{+}]=0.1436M[/tex]
pH = [tex]-log[H^{+}][/tex] = -log(0.1436) = 0.84
