The electrode potential, `E^(@)`, for the reduction of `MnO_(4)^(-)" to "Mn^(2+)` in acidic medium is `+1.51V`. Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for `Zn^(2+), Ag^(+) and Au^(+)` are given as
`Zn_((aq))^(2+)+2e^(-)rarrZn_((s)),E^(@)=-0.762V`
`Ag_((aq))^(+)+e^(-)hArr Ag_((x)), E^(@)=+0.80V`
`Au_((aq))^(+)+e^(-)hArr Au_((s)), E^(@)=+1.69V`

Options

A. Zn and Au
B. Ag and Au
C. Au
D. Zn and Ag

Answer

To determine which metals will be oxidized in the presence of the MnO4⁻ ion, we need to analyze the standard electrode potentials (E°) provided for the reduction of the metals and compare them to the reduction potential of MnO4⁻.

### Step-by-Step Solution:

1. **Identify the Reduction Potentials:**
- The reduction potential for MnO4⁻ to Mn²⁺ in acidic medium is given as E° = +1.51 V.
- The reduction potentials for the metals are:
- Zn²⁺ + 2e⁻ → Zn, E° = -0.762 V
- Ag⁺ + e⁻ → Ag, E° = +0.80 V
- Au⁺ + e⁻ → Au, E° = +1.69 V

2. **Determine the Oxidation Potentials:**
- The oxidation potential is the negative of the reduction potential. Therefore, we can calculate:
- For Zn: E° (oxidation) = -(-0.762 V) = +0.762 V
- For Ag: E° (oxidation) = -(+0.80 V) = -0.80 V
- For Au: E° (oxidation) = -(+1.69 V) = -1.69 V

3. **Compare with MnO4⁻ Reduction Potential:**
- MnO4⁻ can oxidize a species if the reduction potential of that species is lower than +1.51 V.
- Now, we compare the oxidation potentials of the metals:
- Zn: +0.762 V (can be oxidized)
- Ag: +0.80 V (can be oxidized)
- Au: +1.69 V (cannot be oxidized as it is higher than +1.51 V)

4. **Conclusion:**
- The metals that can be oxidized by MnO4⁻ are Zn and Ag, as their oxidation potentials are lower than the reduction potential of MnO4⁻.

### Final Answer:
The metals that will be oxidized are **Zn and Ag**.