LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2024
JEE MAIN 2025 Foundation Course
NEET 2024 Crash Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
404 views
in Chemistry by (37.7k points)

In Nernst equation the constant 0.0592 at 298 K represents the value of :

(a) \(\frac{RT}{nF}\)

(b) \(\frac{RT}{F}\) 

(c) \(\frac{2.303\,RT}{nF}\) 

(d) \(\frac{2.303\,RT}{F}\)

Please log in or register to answer this question.

1 Answer

+1 vote
by (35.9k points)

Option : (d) \(\frac{2.303\,RT}{F}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...