JEE Mains Chemistry Past PapershardNUMERICAL

The potential for the given half cell at 298K is (-)………… × 10-2 V. 2H+ (aq) + 2e- H2 (g) [H+] = 1M, JEE Mains Chemistry Past Papers Chemistry Question

Question

The potential for the given half cell at 298K is (-)………… × 10-2 V. 2H+ (aq) + 2e- H2 (g) [H+] = 1M, H P = 2 atm (Given: 2.303 RT/F = 0.06 V, log2 = 0.3)

Answer: .

💡 Solution & Explanation

2 H o H /H P 0.06 E = E log [H ]    0.06 E = 0.00 log [1]  E = 0.03 0.3 0.9 10 V     

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Mains Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry