Question

Consider the following cell diagram:
Pt(s) | Fe³⁺(aq) , Fe²⁺(aq) || Cl^–(aq) | Cl₂(g) | Pt(s)
The reaction utilized by this cell is

Question 8 options:

	Fe2+(aq) + 2Cl–(aq) --> Fe(s) + Cl2(g)
	Fe(s) + Cl2(g) --> Fe2+(aq) + 2Cl–(aq)
	2Fe3+(aq) + 2Cl–(aq) --> 2Fe2+(aq) + Cl2(g)
	Fe3+(aq) + Cl–(aq) --> Fe2+(aq) + 1/2Cl2(g)
	2Fe2+(aq) + Cl2(g) --> 2Fe3+(aq) + 2Cl–(aq)

Answer 1

Please give me better rattings.

Soluti: Given cell diagram is - Pt(o) [festaq.), Fe2 tag.) Il caq. I C12 (991 Pt 69 Then Half cell reaction can be given as - oleft electrode: - Pt (S) | Festage) , Felaars With half cell reaction : Anode Conidation) :- Feet Fest te o Right electrode :- Cathode (Reduction) ; l'ag. 2 (oo l It C Cl2 + 2e za multiplying eq. O by @ 2 and adding to sequ ☺ iwe I get, - Fernant dé- +241 2 Fest C1263 +26 or la Festaqug + Cl₂ o 2 Fe31991) + 2 diago option (5th is correct with a

Thank you.