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Copper from copper sulphate solution can...

Copper from copper sulphate solution can be displacesd by. (The standard reduction potentials of some electrodes are given below):
`E^(o)(Fe^(2+)//Fe) = - 0.44 V, E^(o) (Zn^(2+)//Zn) = - 0.76 V`
`E^(o) (Cu^(2+)//Cu) = + 0.34 V: E^(o)(Cr^(3+)//Cr) = - 0.74 V`
`E^(o) (H^(+0)/H_(2)) = 0.00V`

A

`H_(2)`

B

Zn oxidizes. Fe

C

Cr

D

All of these

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The correct Answer is:
To solve the question of which metals can displace copper from a copper sulfate solution, we need to analyze the standard reduction potentials given and determine which metals have a higher oxidation potential than copper. ### Step-by-Step Solution: 1. **Identify the Standard Reduction Potentials**: We have the following standard reduction potentials: - \( E^\circ (Fe^{2+}/Fe) = -0.44 \, V \) - \( E^\circ (Zn^{2+}/Zn) = -0.76 \, V \) - \( E^\circ (Cu^{2+}/Cu) = +0.34 \, V \) - \( E^\circ (Cr^{3+}/Cr) = -0.74 \, V \) - \( E^\circ (H^+/H_2) = 0.00 \, V \) 2. **Determine the Oxidation Potentials**: The oxidation potential is the negative of the reduction potential. Therefore, we can calculate the oxidation potentials: - \( E^\circ_{oxidation} (Fe) = +0.44 \, V \) - \( E^\circ_{oxidation} (Zn) = +0.76 \, V \) - \( E^\circ_{oxidation} (Cu) = -0.34 \, V \) - \( E^\circ_{oxidation} (Cr) = +0.74 \, V \) - \( E^\circ_{oxidation} (H_2) = 0.00 \, V \) 3. **Compare Oxidation Potentials**: To displace copper from copper sulfate, we need a metal with a higher oxidation potential than that of copper: - Copper's oxidation potential is \( -0.34 \, V \). - Now we compare: - \( Fe: +0.44 \, V \) (can displace) - \( Zn: +0.76 \, V \) (can displace) - \( Cr: +0.74 \, V \) (can displace) - \( H_2: 0.00 \, V \) (can displace) 4. **Conclusion**: Since all the metals (Fe, Zn, Cr, and H2) have higher oxidation potentials than copper, they can all displace copper from copper sulfate solution. ### Final Answer: Copper from copper sulfate solution can be displaced by **all of these metals** (Fe, Zn, Cr, and H2). ---
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