Given that EH3O/H2∣Pt=0 at 298K.. The pressure of H2 gas would be

Given that $E_{H_{3} O / H_{2} ∣ Pt} = 0 at 298 K .$ . The pressure of $H_{2}$ gas would be

A
$10^{- 7} atm$
B
$10^{- 14} atm$
C
$10^{- 10} atm$
D
$10^{- 12} atm$

Solution:

For water, $[H^{+}] = 10^{- 7} M at 298 K$

$H^{+} + e^{-} \to \frac{1}{2} H_{2}; E = - \frac{R T}{F} \ln \frac{p_{H_{2}}^{1 / 2}}{[H^{+}]}$

Obviously, when E = 0

${(\frac{p_{H_{2}}}{atm})}^{1 / 2} = \frac{[H^{+}]}{{moldm}^{- 3}} = 10^{- 7} or p_{H_{2}} = 10^{- 14} atm$

Related content

SBTET Full Form

What is CSS Scholarship – Know About CSS (Central Sector Scholarship) Scholarship 2023

International Coffee Day

International Day for the Elderly

MPL Full Form – Mobile Premier League

How to Prepare for NTSE 2023-24: Best NTSE Preparation Tips?

RRB NTPC Eligibility 2023

NTSE (National Talent Search Exam) Scholarship Scheme: Details, Amount, Eligibility, Selection Process

Casual Leave Application for Office, Teachers, Government Offices with Format and Samples

NCVT Full Form – National Council for Vocational Training