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.
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is: