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24. november 2013 af Lalaaa (Slettet) - Niveau: A-niveau

Hej alle, jeg sidder med en opgave jeg ikke kan forstå - nogen der kan help ? Consider the cell (Fe(s) lFeSO4(aq, a=0,0250)l lHg2SO4(s)l Hg(l). a)Write the cell reaction. b)Calculate the cell potential, the equilibrium constant for the cell reaction, and DeltaG at 25C?

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Svar #1
02. januar 2015 af Heptan

Fe_(s) | FeSO₄_(aq), a = 0,0250 || Hg₂SO₄_(s) | Hg_(l)

Iron is being oxidized by mercury(I)sulphate. Thus the cell reaction

Fe_(s) + Hg₂SO_{4 (s)} → FeSO₄_(aq) + Hg_(l)

The cell reaction can be thought of as being a combination of the two half reactions with their respective reduction potentials,

Fe_(s) → Fe²⁺_(aq) + 2 e^- E_l^o = - 0,447 V

Hg₂SO₄_(s) + 2 e^- → 2 Hg_(l) + SO₄^2-_(slv) E_r^o = 0,796 V

(sulphate being a spectator ion).

Hence the standard cell potential

E^o_cell = E_r^o - E_l^o = + 1,243 V

and Gibbs free energy

ΔG^o = -zFE^o_cell = - 239,9 kJ/mol

and the equilibrium constant

K^o = exp(-ΔG^o/RT) = 1,07 * 10⁴²

The equilibrium constant is very large, as we would expect for a reaction between a metal being far to the left in the electrochemical and a metal ion being far to the right.

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Svar #2
02. januar 2015 af Heptan

However, the reaction is not under standard conditions, and we must take the reaction quotient into account. The solids and the liquid mercury can all assumed to have an activity close to one, and we need to worry only about iron(II)sulphate. Thus, the reaction quotient is given by

$Q=\frac{1}{0,0250}$

and the cell potential of the reaction can be calculated using Nerns't equation:

$E=E^o - \frac{0,0592}{z}\cdot \log Q =1,20\ V$

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