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A plumber's handbook states that you should not connect a copper pipe directly to a steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use, instead, an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

Short Answer

Expert verified
The spontaneous redox reactions that can cause corrosion between copper and steel pipes are: a) Copper corrodes while iron is oxidized to Fe(2+), with a standard emf value of 0.78V. b) Copper corrodes while iron is oxidized to Fe(3+) first, then reduced to Fe(2+), with a standard emf value of 1.11V. These reactions are spontaneous due to their positive standard emf values and can lead to corrosion. Using an insulating fitting prevents these reactions from occurring, avoiding corrosion.

Step by step solution

01

1. Identify possible redox reactions

In this step, we need to identify the oxidation states of copper and iron and determine the reactions that could occur. Copper can exist in the oxidation states of Cu (0) and Cu (2+), while iron can exist in Fe (0), Fe (2+), and Fe (3+) oxidation states. Possible redox reactions between these metals are: 1. Cu (0) -> Cu (2+) + 2e- (Oxidation of copper) 2. Fe (0) -> Fe (2+) + 2e- (Oxidation of iron) 3. Fe (2+) -> Fe (3+) + e- (Oxidation of iron)
02

2. Determine standard reduction potentials

Next, we will find the standard reduction potentials of these reactions using a standard reduction potential table. 1. Cu (2+) + 2e- -> Cu (0) E° = +0.34V (reduction potential of copper) 2. Fe (2+) + 2e- -> Fe (0) E° = -0.44V (reduction potential of iron) 3. Fe (3+) + e- -> Fe (2+) E° = +0.77V (reduction potential of iron) Now, we will have to reverse the reactions 2 and 3 to obtain their oxidation potentials. 2. Fe (0) -> Fe (2+) + 2e- E° = +0.44V (oxidation potential of iron) 3. Fe (2+) -> Fe (3+) + e- E ° = -0.77V (oxidation potential of iron)
03

3. Calculate standard emf values for possible reactions

The standard emfs of the reactions can be calculated by using the Nernst equation: Ecell = E°cathode - E°anode Possible redox reactions: a) Copper corrodes while iron is oxidized to Fe(2+) Ecell = E°Cu(2+)/Cu - E°Fe/Fe(2+) Ecell = 0.34V - (-0.44V) Ecell = 0.78V b) Copper corrodes while iron is oxidized to Fe(3+) first, then reduced to Fe(2+) Ecell = E°Cu(2+)/Cu - E°Fe(2+)/Fe(3+) Ecell = 0.34V - (-0.77V) Ecell = 1.11V
04

4. Justify the answer with standard emf calculations

Comparing the standard emf values calculated from step 3, both reactions a and b have positive emf values (0.78V and 1.11V) and are therefore spontaneous. However, reaction b has a higher standard emf value, which means it is more spontaneous. This analysis shows that connecting a copper pipe directly to a steel pipe can lead to corrosion due to the spontaneous redox reactions between the metals. The most spontaneous reaction involves the oxidation of copper to Cu(2+) and the oxidation of iron to Fe(3+), followed by reduction to Fe(2+). Using an insulating fitting helps to prevent these reactions from occurring, avoiding corrosion.

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Most popular questions from this chapter

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode, or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

(a) What is a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode?

For the generic reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow\) \(\mathrm{A}^{-}(a q)+\mathrm{B}^{+}(a q)\) for which \(E^{\circ}\) is a positive number, answer the following questions: (a) What is being oxidized, and what is being reduced? (b) If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? (c) Which half-reaction from (b) is higher in potential energy? (d) What is the sign of the free energy change for the reaction? [Sections \(20.4\) and \(20.5]\)

Two important characteristics of voltaic cells are their cell potential and the total charge that they can deliver. Which of these characteristics depends on theamount of reactants in the cell, and which one depends on their concentration?

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., Vol. \(50,1973,61\) ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

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