At 298 K, the standard reduction potential for Cu2+/ Cu electrode is 0.34 V. Given : Ksp Cu(OH)2 = 1 — 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 ×10–20 Take V . F RT . The reduction potential at pH = 14 for the above couple is (-)x × 10–2 V. The value of x is_________________.
Answer: .
💡 Solution & Explanation
pH = 14 pOH = 0 [OH–] = 1 Ksp Cu(OH)2 = [Cu2+] [OH–]2 = 1 ×10–20 [Cu2+] = 10–20 059 . E E Cu | Cu Cu | Cu 2 log ] Cu [ 2 = 0.34 – 059 . log 1020 = – 0.25 | JEE(Main) 2023 | DATE : 13-04-2023 (SHIFT-2) | PAPER-1 | CHEMISTRY PAGE # 9 = –25 × 10–2 V.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes