Answer:
[H⁺] = 1.30X10⁻⁴ M
Explanation:
the problem will be solved by using Nernst's equation, which is :
[tex]E_{cell}=E^{0}_{cell}- \frac{0.0592}{n}logQ[/tex]
In the given equation
n = 2
Q = [tex]=\frac{[Zn^{+2}][p_{H2}]}{[H^{+}]^{2}}[/tex]
Putting values
[tex]0.53= 0.76 - \frac{0.0592}{n}log(\frac{1X1}{[H^{+}]^{2}})[/tex]
on calculating
[H⁺] = 1.30X10⁻⁴ M