Question

12.If the electrode potential of the half cell Pt, H 2 (1 atm) / HCl (aq) at 298 K is -0.12V, the pH of the
acid is
1) 1
13.
2) 1.3 @Rinu
3) 2
4) 3.3

Answer 1

This will help you in finding the answer

Answer 2

Answer:

3) 2

Explanation:

E=E0-0.059/n log(product/reactant)

SRP of H2 electrode is zero

-0.12=0-0.059/1*log(H+/1)

-0.12=0.059*(-log[H+])

0.12/0.059=pH

=2.03

pH=2