The equilibrium constant for the reaction Zn(s) + Sn2β (aq) Zn2+ (aq) + Sn(s) is 1 x 1020 at 298 K. β JEE Mains Chemistry Past Papers Chemistry Question
Question
The equilibrium constant for the reaction Zn(s) + Sn2β (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 10β2 V. (Nearest integer). Given: F RT . = 0.059 V
Answer: .
π‘ Solution & Explanation
Eo = 059 . log Keq. = 059 . log 1020 ο 2 . ο΄ = 0.59V o cell E = o Sn / Sn2 E ο« β o Zn / Zn2 E ο« 0.59 = o Sn / Sn2 E ο« β (β0.76) o Sn / Sn2 E ο« = β0.76 + 0.59 = β0.17 o Sn / Sn E ο« = 0.17 = 17 Γ 10β2 Ans. = 17 | JEE(Main) 2023 | DATE : 29-01-2023 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 9
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