Question

The standard reduction potentials of \( 2H^+/H_2 \), \( Cu^{2+}/Cu \), \( Zn^{2+}/Zn \), and \( NO_3^-/HNO_3 \) are 0.0 V, 0.34 V, -0.76 V, and 0.97 V respectively. Identify the correct statements from the following:
I. \( H^+ \) does not oxidize \( Cu \) to \( Cu^{2+} \). 
II. \( Zn \) reduces \( Cu^{2+} \) to \( Cu \). 
III. \( NO_3^- \) oxidizes \( Cu \) to \( Cu^{2+} \).

• A. I, II only
• B. I, II , III
• C. I, III only
• D. II, III only

Hint:

Oxidation occurs when a species loses electrons, while reduction occurs when a species gains electrons. The species with a higher reduction potential acts as an oxidizing agent.

Answer 1

The Correct Option is B

Step 1: Understanding Standard Reduction Potentials 
\(E^\circ_{\text{red}} (H^+/H_2) = 0.00V, E^\circ_{\text{red}} (Cu^{2+}/Cu) = 0.34V, E^\circ_{\text{red}} (Zn^{2+}/Zn) = -0.76V,\)
\(E^\circ_{\text{red}} (NO_3^-/HNO_3) = 0.97V\)

Step 2: Evaluating Statements - Statement I: 
- \( H^+ \) can only oxidize a metal if its reduction potential is higher than 0.00V.
- \( Cu \) has \( 0.34V \), which is higher than \(H^+\).
- \( H^+ \) cannot oxidize \( Cu \) to \( Cu^{2+} \) . - Statement II: 
- \( Zn \) has a lower reduction potential (-0.76V) than \( Cu^{2+}/Cu (0.34V) \).
- \( Zn \) is a stronger reducing agent, so it can reduce \( Cu^{2+} \) to \( Cu \).
- This statement is correct. - Statement III: 
- \( NO_3^- \) has a higher reduction potential (0.97V) than \( Cu^{2+}/Cu (0.34V) \).
- \( NO_3^- \) can oxidize \( Cu \) to \( Cu^{2+} \).
- This statement is correct.

Answer 2

Given standard reduction potentials:
- \( 2H^+/H_2 = 0.0 \, V \)
- \( Cu^{2+}/Cu = 0.34 \, V \)
- \( Zn^{2+}/Zn = -0.76 \, V \)
- \( NO_3^-/HNO_3 = 0.97 \, V \)

Let's analyze each statement:

I. \( H^+ \) does not oxidize \( Cu \) to \( Cu^{2+} \):
- The oxidation reaction would be: \( Cu \rightarrow Cu^{2+} + 2e^- \)
- The corresponding reduction is \( 2H^+ + 2e^- \rightarrow H_2 \) with \( E^\circ = 0.0 \, V \)
- Since \( Cu^{2+}/Cu \) has a higher reduction potential (0.34 V) than \( H^+/H_2 \), copper is less likely to be oxidized by hydrogen ions.
- Therefore, \( H^+ \) cannot oxidize copper to \( Cu^{2+} \).
Statement I is correct.

II. \( Zn \) reduces \( Cu^{2+} \) to \( Cu \):
- Zinc has a lower reduction potential (-0.76 V) than copper (0.34 V), so zinc metal tends to lose electrons (oxidize) and copper ions gain electrons (reduce).
- The redox reaction is spontaneous:
\( Zn \rightarrow Zn^{2+} + 2e^- \) (oxidation)
\( Cu^{2+} + 2e^- \rightarrow Cu \) (reduction)
Statement II is correct.

III. \( NO_3^- \) oxidizes \( Cu \) to \( Cu^{2+} \):
- \( NO_3^- \) has a higher reduction potential (0.97 V) than \( Cu^{2+}/Cu \) (0.34 V), indicating it is a stronger oxidizing agent.
- Hence, \( NO_3^- \) can oxidize copper metal to \( Cu^{2+} \).
Statement III is correct.

Conclusion:
All three statements I, II, and III are correct.