Question

At one atmosphere pressure, iron (Fe) and nickel (Ni) oxidize as
\(2\text{Fe} + \text{O}_2 ⇒ 2\text{FeO} \Delta G^\circ = -527400 + 128 T \text{ Joules}\)
\(2\text{Ni} + \text{O}_2 ⇒ 2\text{NiO} \Delta G^\circ = -471200 + 172 T \text{ Joules}\)
Identify the correct statement.
Given: Temperature, T is in Kelvin

• A. Fe can reduce NiO at all temperatures
• B. Fe can reduce NiO only above 1000 K
• C. Ni can reduce FeO at all temperatures
• D. Ni can reduce FeO only above 1000 K

Hint:

For reduction reactions of the type M1 + M2O \(⇒\) M1O + M2, the reaction is spontaneous if the Ellingham line for M1 is below the line for M2. To check this mathematically, simply set \(\Delta G^\circ_{M1}<\Delta G^\circ_{M2}\) and solve for T.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In extractive metallurgy, a metal can reduce the oxide of another metal if the Gibbs free energy change (\(\Delta G\)) for the overall reaction is negative. This principle is visualized in Ellingham diagrams, where the oxidation line of the reducing element must lie below the oxidation line of the element being reduced.
Step 2: Key Approach:
We want to check if Fe can reduce NiO. The reaction is: \(2\text{Fe} + 2\text{NiO} ⇒ 2\text{FeO} + 2\text{Ni}\). To find the \(\Delta G^\circ\) for this reaction, we can use Hess's Law. We need to combine the given oxidation reactions.
Reaction 1: \(2\text{Fe} + \text{O}_2 ⇒ 2\text{FeO} \Delta G^\circ_1 = -527400 + 128T\)
Reaction 2: \(2\text{Ni} + \text{O}_2 ⇒ 2\text{NiO} \Delta G^\circ_2 = -471200 + 172T\)
We need the reduction of NiO, so we reverse Reaction 2: Reverse Reaction 2: \(2\text{NiO} ⇒ 2\text{Ni} + \text{O}_2 \Delta G^\circ = -\Delta G^\circ_2 = -(-471200 + 172T)\)
Now, we add Reaction 1 and the reversed Reaction 2:
\[ (2\text{Fe} + \text{O}_2) + (2\text{NiO}) ⇒ (2\text{FeO}) + (2\text{Ni} + \text{O}_2) \] The overall reaction is: \(2\text{Fe} + 2\text{NiO} ⇒ 2\text{FeO} + 2\text{Ni}\). The Gibbs free energy for this reaction is \(\Delta G^\circ_{rxn} = \Delta G^\circ_1 - \Delta G^\circ_2\). The reaction is spontaneous (i.e., Fe can reduce NiO) if \(\Delta G^\circ_{rxn}<0\), which means \(\Delta G^\circ_1<\Delta G^\circ_2\).
Step 3: Detailed Calculation:
We set up the inequality \(\Delta G^\circ_1<\Delta G^\circ_2\): \[ -527400 + 128T<-471200 + 172T \] Now, solve for T. \[ -527400 + 471200<172T - 128T \] \[ -56200<44T \] \[ T>\frac{-56200}{44} \] \[ T>-1277.27 \text{ K} \] Since the temperature T must be positive in the Kelvin scale (\(T \ge 0\)), this inequality is true for all physically possible temperatures.
Step 4: Final Answer:
The condition for Fe to reduce NiO is always met. Therefore, Fe can reduce NiO at all temperatures.
Step 5: Why This is Correct:
The calculation shows that the Gibbs free energy for the oxidation of iron is always more negative than that for the oxidation of nickel, for any positive temperature. This means the Ellingham line for Fe is always below the line for Ni, making Fe a better reducing agent for NiO at all temperatures.