Question

At 298 K, the standard electrode potentials of Cu²⁺/Cu, Zn²⁺/Zn, Fe²⁺/Fe and Ag⁺/Ag are 0.34 V, - 0.76 V, - 0.44 V and 0.80 V, respectively.
On the basis of standard electrode potential, predict which of the following reaction cannot occur?

• A. CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)
• B. CuSO₄(aq) + Fe(s) → FeSO₄(aq) + Cu(s)
• C. FeSO₄(aq) + Zn(s) → ZnSO₄(aq) + Fe(s)
• D. 2CuSO₄(aq) + 2Ag(s) → 2Cu(s) + Ag₂SO₄(aq)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A redox reaction is spontaneous if the standard cell potential ($E^\circ_{cell}$) is positive.
$E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ (using reduction potentials).
A metal with a lower reduction potential can reduce (displace) a metal ion with a higher reduction potential from its salt solution.
Step 2: Detailed Explanation:
Let's check the $E^\circ_{cell}$ for each reaction:
1. Zn displacing Cu: $E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Zn^{2+}/Zn} = 0.34 - (-0.76) = +1.10$ V. (Spontaneous)
2. Fe displacing Cu: $E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Fe^{2+}/Fe} = 0.34 - (-0.44) = +0.78$ V. (Spontaneous)
3. Zn displacing Fe: $E^\circ_{cell} = E^\circ_{Fe^{2+}/Fe} - E^\circ_{Zn^{2+}/Zn} = -0.44 - (-0.76) = +0.32$ V. (Spontaneous)
4. Ag displacing Cu: Here, $Ag$ is oxidized and $Cu^{2+}$ is reduced.
$E^\circ_{cell} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Ag^+/Ag}$
$E^\circ_{cell} = 0.34 - 0.80 = -0.46$ V.
Since $E^\circ_{cell}$ is negative, this reaction is non-spontaneous and cannot occur.
Step 3: Final Answer:
Reaction (4) cannot occur because Silver has a higher reduction potential than Copper and cannot reduce $Cu^{2+}$ ions.