Consider the cell Pt(s) | H2(g) (1 atm) | H+ (aq, [H+] = 1) || Fe3+ (aq), Fe2+ (aq) | Pt(s) Given : — JEE Mains Chemistry Past Papers Chemistry Question
Question
Consider the cell Pt(s) | H2(g) (1 atm) | H+ (aq, [H+] = 1) || Fe3+ (aq), Fe2+ (aq) | Pt(s) Given : E° 3 Fe / Fe E = 0.771 V and H 1 / H E = 0 V , T = 298 K If the potential of the cell is 0.712 V, the ratio of concentration of Fe2+ to Fe3+ is _________ (Nearest integer)
Answer: .
💡 Solution & Explanation
Anode 1 H2(g) H+ (aq) + e– Cathode Fe3+ + e– Fe2+ –––––––––––––––––––––––––––––– Overall 1 H2 + Fe3+ 1 n H+ + Fe2+ Ecell = E0cell – 059 .0 log 1 H 2 ] P [ ] H [ ] Fe [ ] Fe [ 0.712 = 0.771 – 0.059 log ] Fe [ ] Fe [ 2 log ] Fe [ ] Fe [ 2 = 1 So ] Fe [ ] Fe [ 2 = 10
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