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xeze [42]
3 years ago
11

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) Writ

e 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?
Chemistry
1 answer:
34kurt3 years ago
4 0

<u>Answer:</u>

<u>For a:</u> The balanced chemical equation is written below.

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

<u>For c:</u> Zinc will not protect iron pipe from corrosion.

<u>Explanation:</u>

  • <u>For a:</u>

The given half reaction follows:

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

<u>Oxidation half reaction:</u>  Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V

<u>Reduction half reaction:</u>  Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V

The balanced chemical equation follows:

Fe+Zn^{2+}\rightarrow Fe^{2+}+Zn

  • <u>For b:</u>

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

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

E^o_{cell}=E^o_{cathode}-E^o_{anode}

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

Calculating the E^o_{cell} using above equation, we get:

E^o_{cell}=-0.441-(-0.763)=0.322V

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.

  • <u>For c:</u>

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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