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)?

Please show the work when calculating these values. thank you!!

Explanation / Answer

a)

we know that

oxidation takes place at anode

so

Pb/Pb+2 is the oxidation half reaction

oxidation half reaction : anode

Pb (s) ---> Pb+2 + 2e-

b)

reduction takes place at cathode

Ag+/Ag is reduction half reaction

Ag+ + e- ---> Ag (s)

Pb (s) ----> Pb+2 + 2e-

2Ag+ + 2e- --> 2 Ag (s)

c)

so n =2 as two electrons are transferred

d)

now

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Ag+/Ag - Eo Pb+2/Pb

Eo cell = 0.8 + 0.13

Eo cell = 0.93 V

e)

the overall cell reaction is

2Ag+ + Pb (s) ---> 2 Ag (s) + Pb+2

Q = [Pb+2] / [Ag+]^2

f)

Q = [2] / [0.004]^2

Q = 125000

now

according to nernst equation

E = Eo - (0.05916/n) log Q

E = 0.93 - ( 0.05916/2) log 125000

E = 0.779

so

E cell value is 0.779 V

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