1. Consider the cell represented by this diagram: What is the emf (E cell) for t

Question

1. Consider the cell represented by this diagram: What is the emf (E cell) for the reaction as written at 298K? Pb(s) / Pb2+(2.0M) // Ag+(0.0040M) / Ag(s) Must show work!

a. What is oxidation half reaction?

Now show oxidation half reaction with electrons.

b. What is reduction half reaction?

Now show reduction half reaction with electrons.

Now multiply reduction half reaction by appropriate factor so electrons are the same.

c. What is n?

d. What is Eocell in volts?

e. What is expression for Q in terms of reagents?

f. What is Q numerically? ___________________

g. What is the emf (E cell) for the reaction as written at 298K (in volts)?

Explanation / Answer

A) Pb(s) --------->   Pb+2    + 2e

b) Ag+ +e -------> Ag

c)    n = 2

d) Eo cell = Eo Pb/Pb+2 + EoAg+/Ag = 0.19 +0.8 = 0.99V

e) Q = [Pb+2]/[Ag+]2

f)      Q =   2/ (0.004)2 = 12500

g)   Ecell = Eo cell - 0.0591/2log([Pb+2]/[Ag+]2) =0.99 -0.0591/2log(12500)

                 = 0.99 -(0.0591x4.09)/2 = 0.869 volt

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