Suppose that the concentrations of NaF and KCI were each 0.10 M in the cell: Pb(

Question

Suppose that the concentrations of NaF and KCI were each 0.10 M in the cell: Pb(s) | pbF_2(s) | F (aq) ||CI' (aq) | Afci (s) | Ag (s) (a) Using the half reactions given below and the nermst equation, calculate the cell voltage: 2AfCI(s) + 2e 2Ag(s) + 2CR (aq) E degree = 0.222 V PbF_2 (s) + 2e Pb (s) + 2F (aq) E degree = -0.350 V (b) Which direction do the electrons flow in this cel? (c) Now calculate the cell votage using the following half reactions (use the Nernst equation): (There shold be reasonable agreement between your voltage from part(a) and the voltage you calculate in this part)

Explanation / Answer

The overall balaned reaction is :-

2AgCl(s) + Pb(s) + 2F-(aq) ----> PbF2(s) + Ag(s) + 2Cl-(aq)

E0cell = 0.222- -(0.35) = 0.772 V

Now, using Nernst equation :-

Ecell = E0cell - (0.059/2)*log{0.1/0.1} = 0.772 V

The electron flow is from anode to cathode i.e. from Pb electrode to Ag electrode.

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