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The E^(o) for half cells Fe//Fe^(2+) and...

The `E^(o)` for half cells `Fe//Fe^(2+) and Cu//Cu^(2+)` are `–0.44` V and ` +0.32 `V respectively. Then which of these is true?

A

`Cu^(2+)` oxidises Fe

B

`Cu^(2+)` oxidises `Fe^(2+)`

C

Cu oxidises `Fe^(2+)`

D

Cu reduces `Fe^(2+)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the standard electrode potentials (E°) of the half-cells given for iron (Fe) and copper (Cu). ### Step-by-Step Solution: 1. **Identify the Given Values**: - For the half-cell reaction of iron: \[ \text{Fe} \leftrightarrow \text{Fe}^{2+} + 2e^{-} \quad (E^{\circ} = -0.44 \, V) \] - For the half-cell reaction of copper: \[ \text{Cu}^{2+} + 2e^{-} \leftrightarrow \text{Cu} \quad (E^{\circ} = +0.32 \, V) \] 2. **Determine the Oxidation Potentials**: - The oxidation potential for iron (Fe to Fe²⁺) is given as: \[ E^{\circ}_{\text{oxidation, Fe}} = -0.44 \, V \] - The oxidation potential for copper (Cu to Cu²⁺) can be derived from the reduction potential: \[ E^{\circ}_{\text{oxidation, Cu}} = -E^{\circ}_{\text{reduction, Cu}} = -0.32 \, V \] 3. **Compare the Oxidation Potentials**: - We compare the oxidation potentials: \[ E^{\circ}_{\text{oxidation, Fe}} = -0.44 \, V \quad \text{and} \quad E^{\circ}_{\text{oxidation, Cu}} = -0.32 \, V \] - Since -0.32 V is greater than -0.44 V, this means that copper has a higher oxidation potential than iron. 4. **Determine the Oxidizing and Reducing Agents**: - The species with the higher oxidation potential (Cu²⁺) will act as a better oxidizing agent, while the species with the lower oxidation potential (Fe) will act as a reducing agent. - Therefore, Cu²⁺ can oxidize Fe to Fe²⁺. 5. **Conclusion**: - The correct statement is that copper ions (Cu²⁺) can oxidize iron (Fe) to iron ions (Fe²⁺). ### Final Answer: The correct option is that **Cu²⁺ oxidizes Fe**.
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Knowledge Check

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