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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 reaction that might cause corrosion when a copper pipe is connected directly to a steel pipe is given by: \( Cu^{2+} + Fe \rightarrow Cu + Fe^{2+} \) The standard emf calculation shows that the reaction is spontaneous with \( E°(cell) = 0.78V \). This electrochemical reaction would cause corrosion, and to prevent it, an insulating fitting should be used to connect the copper and steel pipes.

Step by step solution

01

List the standard reduction potentials for copper and iron

First, let's look at the standard reduction potentials for copper and iron: Cu^(2+) + 2e^(-) -> Cu E° = +0.34V Fe^(2+) + 2e^(-) -> Fe E° = -0.44V Fe^(3+) + 3e^(-) -> Fe E° = -0.04V
02

Determine possible redox reactions

Now, let's determine the possible redox reactions between copper and iron. Since we want the reactions to be spontaneous, we'll look for combinations where the reduction potential of copper is greater than that of iron: Cu^(2+) + 2e^(-) -> Cu E° = +0.34V Fe^(2+) + 2e^(-) -> Fe E° = -0.44V Fe^(3+) + 3e^(-) -> Fe E° = -0.04V A possible redox reaction would be between Cu^(2+), acting as the oxidizing agent, and Fe, acting as the reducing agent.
03

Write the balanced overall redox reaction

The balanced overall redox reaction would be: Cu^(2+) + Fe -> Cu + Fe^(2+)
04

Calculate the standard emf

Now, let's calculate the standard emf for this redox reaction using the standard reduction potentials: E°(cell) = E°(cathode) - E°(anode) E°(cell) = 0.34V - (-0.44V) = 0.78V Since E°(cell) > 0, the reaction is spontaneous.
05

Conclusion

The spontaneous redox reaction between a copper pipe and a steel pipe is given by: Cu^(2+) + Fe -> Cu + Fe^(2+) This reaction would cause corrosion due to the electrochemical reactions between the two metals. The standard emf calculation shows that the reaction is spontaneous with E°(cell) = 0.78V. To prevent this corrosion, an insulating fitting should be used to connect the copper and steel pipes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard EMF Calculation
When we talk about standard emf calculation, we're essentially determining the potential difference generated by a redox reaction, under standard conditions. **Standard conditions** mean:
  • Each solution concentration is 1 M
  • Pressure is at 1 atm for any gases involved
  • Temperature is typically at 25°C or 298 K
To calculate the standard emf (E^{°}_{cell}), we use the standard reduction potentials of the substances involved. We subtract the anode (site of oxidation) potential from the cathode (site of reduction) potential:
\[E^{°}_{ ext{cell}} = E^{°}_{ ext{cathode}} - E^{°}_{ ext{anode}}.\]The standard reduction potential for copper is +0.34V, which is higher than iron's -0.44V.
This difference indicates that copper ions are more easily reduced compared to iron ions. When the standard emf is positive, as in our exercise, it means the redox reaction is spontaneous.
Copper and Iron Reaction
A common issue that arises when connecting copper and iron pipes is their interaction through electrochemical reactions. Copper and iron, when in contact, create a galvanic cell. In this scenario:
  • Copper acts as the cathode (undergoes reduction)
  • Iron acts as the anode (undergoes oxidation)

The relevant redox reaction is:
\[ ext{Cu}^{2+} + ext{Fe} ightarrow ext{Cu} + ext{Fe}^{2+}\]Here, copper ions gain electrons (reduction), and iron metal loses electrons (oxidation), forming iron ions. This transfer of electrons causes iron to corrode.
Since the emf calculation resulted in a positive value (+0.78V), it confirms the spontaneous nature of this reaction, leading to iron oxidation.
Galvanic Corrosion Prevention
To prevent galvanic corrosion when connecting copper to steel pipes, we need to consider the electrochemical interactions at play. Galvanic corrosion happens because the two metals facilitate a redox reaction.
Strategies to mitigate corrosion:
  • Use an **insulating fitting** to separate the copper and iron, preventing direct contact.
  • Apply **protective coatings** to the surfaces to minimize exposure.
  • Ensure **proper material selection**, avoiding pairings prone to galvanic interactions.
By breaking the electrical path, corrosion can be prevented efficiently. The insulating fitting recommendation from the plumber's handbook effectively deters the spontaneous reactions, thereby preserving the integrity of the piping system.

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Most popular questions from this chapter

Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{aligned} &\mathrm{HgO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ &\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{aligned} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

(a) A voltaic cell is constructed with all reactants and products in their standard states. Will this condition hold as the cell operates? Explain. (b) Can the Nernst equation be used at temperatures other than room temperature? Explain. (c) What happens to the emf of a cell if the concentrations of the products are increased?

(a) Write the reactions for the discharge and charge of a nickel-cadmium rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Consider the half reaction \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\) (a) Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of \(\mathrm{Ag}^{+} ?\) (b) What is the value of \(E_{\text {red }}\) when \(\log \left[\mathrm{Ag}^{+}\right]=0 ?[\) Section \(20.6]\)

Using data in Appendix \(\mathrm{E}\), calculate the standard emf for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathbf{F}^{-}(a q)\) (b) \(\mathrm{Cu}^{2+}(a q)+\mathrm{Ca}(\mathrm{s}) \longrightarrow \mathrm{Cu}(s)+\mathrm{Ca}^{2+}(a q)\) (c) \(3 \mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q)\) (d) \(\mathrm{Hg}_{2}{ }^{2+}(a q)+2 \mathrm{Cu}^{+}(a q) \longrightarrow 2 \mathrm{Hg}(l)+2 \mathrm{Cu}^{2+}(a q)\)

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