Problem 87 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The spontaneous redox reaction that might cause corrosion is between copper (present in brass) and zinc, with a standard emf of +1.10 V \( (Cu^{2+}(aq) + Zn(s) \rightarrow Cu(s) + Zn^{2+}(aq)) \). This electrochemical reaction can occur when connecting a brass pipe directly to a galvanized steel pipe. Using an insulating fitting prevents direct contact between the metals, avoiding the corrosion process.

Step 1: Identify the half-reactions

First, we need to identify the possible half-reactions for the metals involved: copper, zinc, and iron (the main component of steel). We will consider the following half-reactions: 1. Copper (Cu): \( Cu^{2+}(aq) + 2e^- ⟶ Cu(s) \) 2. Zinc (Zn): \( Zn^{2+}(aq) + 2e^- ⟶ Zn(s) \) 3. Iron (Fe): \( Fe^{2+}(aq) + 2e^- ⟶ Fe(s) \)

Step 2: Determine the standard reduction potentials

Next, we should look up the standard reduction potentials (Eº) for each half-reaction. These values can be found in a standard reduction potentials table. We have: 1. Copper (Cu): \( E^\circ_{Cu} = +0.34 \, V \) 2. Zinc (Zn): \( E^\circ_{Zn} = -0.76 \, V \) 3. Iron (Fe): \( E^\circ_{Fe} = -0.44 \, V \)

Step 3: Determine the possible redox reactions

Now, we will examine the possible combinations of half-reactions between these metals, considering the two half-reactions with the higher standard reduction potentials to be reduction reactions and the one with the lower standard reduction potential to be the oxidation reaction: 1. Copper and Zinc: Cu reduction and Zn oxidation 2. Copper and Iron: Cu reduction and Fe oxidation 3. Zinc and Iron: Zn reduction and Fe oxidation We can now calculate the standard emf for each possible reaction to determine which one is spontaneous.

Step 4: Calculate the standard emf for each reaction

The standard emf is given by the difference in standard reduction potentials, represented by: \( E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \) For each possible reaction: 1. Copper and Zinc: \( E^\circ_{cell} = E_{Cu}^\circ - E_{Zn}^\circ = (+0.34 \, V) - (-0.76 \, V) = +1.10 \, V \) 2. Copper and Iron: \( E^\circ_{cell} = E_{Cu}^\circ - E_{Fe}^\circ = (+0.34 \, V) - (-0.44 \, V) = +0.78 \, V \) 3. Zinc and Iron: \( E^\circ_{cell} = E_{Zn}^\circ - E_{Fe}^\circ = (-0.76 \, V) - (-0.44 \, V) = -0.32 \, V \)

Step 5: Determine the spontaneous redox reaction

From the calculated standard emf values, we can see that the reactions with positive values are spontaneous: 1. Copper and Zinc: \( E^\circ_{cell} = +1.10 \, V \) 2. Copper and Iron: \( E^\circ_{cell} = +0.78 \, V \) The reaction with the highest standard emf is the most spontaneous, in this case, the reaction between copper and zinc: \( Cu^{2+}(aq) + Zn(s) \rightarrow Cu(s) + Zn^{2+}(aq) \) So, the spontaneous redox reaction that might cause corrosion is between copper (present in brass) and zinc. This is why it's not recommended to connect a brass pipe directly to a galvanized steel pipe. Using an insulating fitting prevents direct contact between the two metals, thus avoiding the corrosion process.