Pt(s)|Fe2+(aq)|Fe3+(aq)||Ag+(aq)|Ag(s) Why when we are figuring out the oxidatio

Question

Pt(s)|Fe2+(aq)|Fe3+(aq)||Ag+(aq)|Ag(s)

Why when we are figuring out the oxidation, where did the H2O come from? because I would of started my left side of the equation with O2 changint to H^+. So is there something that iam missing fundamentally? And what does Pt(s) stand for? what role does it play in this ?

(1) oxidation occurs at the anode: 2 H20 (1) => 02(g) + 4 H+(aq) + 4 e Reduction occurs at the cathode: 02(g) + 2 H20 (1) + 4 e- 4 OH-(aq) => (2) Eo (cellEo(o2/oH-) - Eo (o2/H20) = 0.40-1.23 = -0.83 V

Explanation / Answer

here Pt is inert electode . the above solution is wrong.why you start with O2 . i do not understand . it is clearly given both half -cells

correct way to sove this problem

left side oxidation:

Fe+2 -----------------> Fe+3 + e- , Eo = 0.77 V

right side reduction :

Ag+ + e- --------------> Ag , Eo = 0.80 V

Eocell = Eo red - Eo oxd

= 0.80 - 0.77

= 0.03 V

Eocell  = 0.03 V

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