[cliverlong]

Hi,

   The standard method for metal M seems to be place M in a salt solution of same metal, say M²⁺SO₄^2-

   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

[DevaDevil]

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

1M Cr₂O₇^2- in 1M acidic solution (convenient because of the equilibrium with CrO₄^2-, 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