The equilibrium constant for the reaction Zn(s) + Sn2– (aq) Zn2+ (aq) + Sn(s) is 1 x 1020 at 298 K. — JEE Mains Chemistry Past Papers Chemistry Question
Question
The equilibrium constant for the reaction Zn(s) + Sn2– (aq) Zn2+ (aq) + Sn(s) is 1 x 1020 at 298 K. The magnitude of standard electrode potential of Sn/Sn2+ if Zn / E Zn = -0.76 V is_____________ x 10–2 V. (Nearest integer). Given: F RT . = 0.059 V
Answer: .
💡 Solution & Explanation
Eo = 059 . log Keq. = 059 . log 1020 2 . = 0.59V o cell E = o Sn / Sn2 E – o Zn / Zn2 E 0.59 = o Sn / Sn2 E – (–0.76) o Sn / Sn2 E = –0.76 + 0.59 = –0.17 o Sn / Sn E = 0.17 = 17 × 10–2 Ans. = 17 | JEE(Main) 2023 | DATE : 29-01-2023 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 9
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes