March 28, 2024, 09:21:42 PM
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Topic: Determining electrode potential for non-metals and electropositives  (Read 3091 times)

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Offline cliverlong

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Hi,

   The standard method for metal M seems to be place M in a salt solution of same metal, say M2+SO42-

   Then measure pd when connected to hydrogen cell (or adjust pd when using calomel cell)

Now consider

1/2 Cl2(g)  + e- <> Cl-(aq) E= +1.36V  (1)

How do we measure E?

since

H+(aq) + e- <> 1/2 H2(g) E= 0.0 V  (2)

Since the second equation is "more negative" that reaction flows right to left and hydrogen is oxidized to H+ by loss of electron

Symmetrically, the chlorine gas is reduced by gain of electron to chloride ion Cl-

So is the following apparatus OK?

Reaction (2) : Standard hydrogen cell and bubble hydrogen over platinum electrode submerged in 1 molar hydrochloric acid

Reaction (1) : Hydrogen cell apparatus but bubble chlorine gas over platinum electrode submerged in 1 molar NaCl solution ?



How would apparatus be set up to determine E for the following two monsters?

Cr2O7(aq) + 14H+(aq) + 6e- <> 2Cr3++(aq) + 7H2O(l)  E = +1.33V

MnO4-(aq) + 8H+(aq) + 5e- <> Mn2+(aq) + 4H2O(l)      E = +1.52V


Ta,

Clive

Offline DevaDevil

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yes, you bubble chlorine gas over an inert electrode (Pt does well) in a solution with its ion.

1M Cr2O72- in 1M acidic solution (convenient because of the equilibrium with CrO42-, acid pushes the equilibrium towards dichromate) with 1 M Cr3+, connected to a hydrogen cell.

1M MnO4- and 1 M Mn2+ in a 1M acidic solution with a Au electrode connected to a hydrogen cell

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