Consider the cell Pt(s)|H2 (g, 1 atm)|H+ (aq,1 M)|| Fe3+(aq), Fe2+(aq)|Pt(s) When the potential of t — JEE Mains Chemistry Past Papers Chemistry Question
Question
Consider the cell Pt(s)|H2 (g, 1 atm)|H+ (aq,1 M)|| Fe3+(aq), Fe2+(aq)|Pt(s) When the potential of the cell is 0.712 V at 298 K, the ratio [Fe2+ [Fe3+] is ______ . (Nearest integer) Given: Fe3+ + e– ⇌ Fe2+, E0Fe3+, Fe2+|Pt = 0.771 V . F RT .
💡 Solution & Explanation
Pt(s)|H2 (g, 1 atm)|H+ (aq,1 M)|| Fe3+(aq), Fe2+(aq)|Pt(s) At anode: H2 2H+ + 2e– At cathode: (Fe3+(aq) + e– Fe2+(aq) × 2 overall cell reaction : H2(g) + 2Fe3+ 2H+(aq) + 2Fe2+(aq) Eºcell = Eºcathode – EºAnode = 0.771 – 0 = 0.771 V Eºcell = Eºcell – log n . 2 H 2 ] Fe .[ P ] H [] Fe [ | JEE(Main) 2023 | DATE : 30-01-2023 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 9 0.712 = 0.771 – log 06 . 2 2 ] Fe [ ] Fe [ log ] Fe [ ] Fe [ 2 = 06 . 712 . 771 . ] Fe [ ] Fe [ 2