Given the E0 values of the following two half-reactions: Zn  Zn2+ + 2e- E0 = 0.763 volt Fe  Fe2+ + 2e- E0 = 0.441 volt a) Write a balanced complete oxidation-reduction reaction? b) Explain whether the corrosion of an iron pipe (i.e., Fe  Fe2+) in the presence of Zn/Zn2+ is possible or not (thermodynamically)? c) Explain whether or not Zn will protect the corrosion of iron pipe if metallic Zn is in contact with the iron pipe?

Respuesta :

Answer:

For a: The balanced chemical equation is written below.

For b: The corrosion of iron pipe will take place in the presence of zinc.

For c: Zinc will not protect iron pipe from corrosion.

Explanation:

  • For a:

The given half reaction follows:

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, zinc will undergo reduction reaction will get reduced.

Oxidation half reaction:  [tex]Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V[/tex]

Reduction half reaction:  [tex]Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V[/tex]

The balanced chemical equation follows:

[tex]Fe+Zn^{2+}\rightarrow Fe^{2+}+Zn[/tex]

  • For b:

For a reaction to be spontaneous (thermodynamically feasible) , the standard electrode potential must be positive.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the [tex]E^o_{cell}[/tex] using above equation, we get:

[tex]E^o_{cell}=-0.441-(-0.763)=0.322V[/tex]

As, the EMF is coming out to be positive, the reaction will be thermodynamically feasible and corrosion of iron pipe will take place in the presence of zinc.

  • For c:

As, the EMF of the cell is positive, the zinc will not protect the iron pipe from corrosion and the reaction will take place.

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