(a -5 points) You are trying to protect some lead pipes anode (something else th

Question

(a -5 points) You are trying to protect some lead pipes anode (something else that is oxidized). U this application. Write the full electrochemical equation by combining the two half-cell reaco 2. (35 points) (a - 5 points) You are trying to protect some lead pipes from corrosion by including a sacrificial sing the following table, select a suitable material for Auch + 3e- Au + 4C1- Pd2+ + 2e- Pd Ag+ + e- Ag Cu2+ + 2e- Cu Pb2 + + 2e- Pb Fe2+ +2e- Fe Zn2+ + 2e- Zn V"= 1.0V 0.987V VM=0.SV Vo0.34v V"=-0.13V V--0,45V Vo=-0. 76V (b - 15 points) At room temperature a galvanic cell with Pd and Au is constructed with 1M concentrations each of Pd2 and AuCla ions respectively. Leaving AuCli at 1M, to what must we adjust the concentration of Pd2* ions in order to reverse the spontaneity of this reaction? Report answer in molarity of Pd2* (c - 15 points) Consider the reaction for converting iron oxide to iron metal:

Explanation / Answer

Answer –

a) We are given, standard reduction potential table and we need to write the two half reaction that are showing the protect the lead pipe-

For the protect the lead pipe we need to gets reduction on the lead pipe and need to choose such metal that can acts as oxidized. So we need to choose the metal which has less standard reduction potential than lead, so Zinc metal is very suitable and acts a strong reducing agent.

So two half reaction as follow –

Zn(s) ------> Zn2+(aq)+ 2e- Eooxd = +0.76 V

Pb2+(aq)+2e- ------> Pb(s) Eored = -0.13 V

When we combine this two half reaction then

Zn(s) + Pb2+(aq) ------> Zn2+(aq) + Pb(s) Eocell = 0.63 V

b) The galvanic cell between Pd and Au constructed with 1 M each Pb2+ and AuCl4-. In this cell the AuCl4- acts as the oxidizing agent and gets reduced to Au and Pb gets oxidized and act as reducing agent. So two half reaction are as bellow –

3 Pb(s) ------> 3Pb2+(aq)+ 6e- Eooxd = +0.13 V

2AuCl4- (aq)+6e- ------>2Au(s) + 8Cl- Eored = 1.0 V

So the overall reaction

3 Pb(s)+ 2AuCl4- (aq) ------> 3Pb2+(aq)+ 2Au(s) + 8Cl- Eocell = 1.13 V

So the Pb2+ is in the product side and if we need to reverse the spontaneity then we need to decease the concentration 2/3 so the reverse the spontaneity,

So molarity of Pb2+(aq) = 0.67 M

c) We are given the reaction –

FeO(s) + CO(g) -------> Fe(s) + CO2(g)

And we need to calculate the standard enthalpy change using the given reactions with standard change enthalpies. We need to use Hess’s law

We need to arrange the given three equation in such way that when we added the these three equation then we need to get the above reaction

Reactions are as bellow –

3 Fe2O3(s) + CO(g) -----> 2 Fe3O4(s) + CO2(g) Ho = -47.0 kJ …..1

Fe2O3(s) + 3 CO(g) -----> 2 Fe(s) + 3CO2(g)    Ho = -25.0 kJ …..2

Fe3O4(s) + CO(g) -----> 3FeO(s) + CO2(g)        Ho = 19.0 kJ ……3

Now we need FeO(s) in the reactant side and Fe(s) in the product side so we need to reverse the reaction number 3 and 1

2 Fe3O4(s) + CO2(g) ----> 3 Fe2O3(s) + CO(g) Ho = 47.0 kJ …..4

Fe2O3(s) + 3 CO(g) -----> 2 Fe(s) + 3CO2(g)    Ho = -25.0 kJ…..5

3FeO(s) + CO2(g) -----> Fe3O4(s) + CO(g)      Ho = -19.0 kJ…..6

Now we need to multiply by 3 to equation number 5 and 2 by to equation number 6

2 Fe3O4(s) + CO2(g) ----> 3 Fe2O3(s) + CO(g) Ho = 47.0 kJ …..7

3 Fe2O3(s) + 9 CO(g) -----> 6 Fe(s) + 9CO2(g)    Ho = -75.0 kJ…..8

6FeO(s) + 2CO2(g) -----> 2Fe3O4(s) + 2CO(g)      Ho = -38.0 kJ…..9

Now we need to added the equation number 7 , 8 and 9

2 Fe3O4(s) + CO2(g) ----> 3 Fe2O3(s) + CO(g) Ho = 47.0 kJ …..7

3 Fe2O3(s) + 9 CO(g) -----> 6 Fe(s) + 9CO2(g)   Ho = -75.0 kJ…..8

6FeO(s) + 2CO2(g) -----> 2Fe3O4(s) + 2CO(g) Ho = -38.0 kJ…..9

_____________________________________________________________

6 FeO(s) + 6 CO(g) -------> 6 Fe(s) + 6 CO2(g)   Ho = -66 kJ …..10

Now we need to divide by 6 to equation number 10

So it is , FeO(s) + CO(g) -------> Fe(s) + CO2(g) Ho = -11 kJ

So standard enthalpy of the given reaction is -11 kJ

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