At 298 K, the standard reduction potential for Cu2+/ Cu electrode is 0.34 V. Given : Ksp Cu(OH)2 = 1 β JEE Mains Chemistry Past Papers Chemistry Question
Question
At 298 K, the standard reduction potential for Cu2+/ Cu electrode is 0.34 V. Given : Ksp Cu(OH)2 = 1 Γ10β20 Take V . F RT . The reduction potential at pH = 14 for the above couple is (-)x Γ 10β2 V. The value of x is_________________.
Answer: .
π‘ Solution & Explanation
pH = 14 ο pOH = 0 [OHβ] = 1 Ksp Cu(OH)2 = [Cu2+] [OHβ]2 = 1 Γ10β20 [Cu2+] = 10β20 059 . E E Cu | Cu Cu | Cu 2 ο ο½ ο« ο« log ] Cu [ 2ο« = 0.34 β 059 . log 1020 = β 0.25 | JEE(Main) 2023 | DATE : 13-04-2023 (SHIFT-2) | PAPER-1 | CHEMISTRY PAGE # 9 = β25 Γ 10β2 V.
π¬Ask on WhatsApp β
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp β gets answered in minutes