Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Q.

E⁰ for $F_{2} + 2 \bar{e} \to 2 \bar{F}$ is 2.8
E⁰ for $\frac{1}{2} F_{2} + \bar{e} \to \bar{F}$ is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

a

1.4 V

b

– 2.8 V

c

2.8 V

d

– 1.4 V

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Electrode potentials are intensive properties .

$E = {E^o} - \frac{{0.0591}}{1}\log \frac{{\left[ {{F^ - }} \right]}}{\left[ {{F_2}} \right]^{\frac{1}{2}}}$

$= {E^o} = 2.8V$

Since [F^-] = 1M and pressure of F₂=1 atm

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!

E0 for F2 +2e- →2F- is 2.8 E0 for 12F2 +e- →F- is