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Standard electrode potential data are us...

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below:
`MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V`
`Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V`
`Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V`
`CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V`
Identify the only correct statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`

A

`MnO_(4)^(-)` can be used in aqueous HCI

B

`Cr_2O_(7)^(2-)` can be used in aqueous HCI

C

`Mn_4^(-)` can be used in aqueous `H_2SO_4`

D

`Cr_2O_(7)^(2-)` can be used in aqueous `H_2SO_4`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem regarding the quantitative estimation of aqueous `Fe(NO3)2`, we need to analyze the provided half-cell reactions and their standard electrode potentials. Here’s a step-by-step solution: ### Step 1: Identify the oxidation state of iron in `Fe(NO3)2` `Fe(NO3)2` contains iron in the +2 oxidation state (Fe²⁺). For quantitative estimation, we need to oxidize Fe²⁺ to Fe³⁺. **Hint:** Look for the oxidation states of the elements in the compound to understand the redox process. ### Step 2: Determine the suitable oxidizing agents From the provided half-reactions, we can identify potential oxidizing agents: - `MnO4^−` (E° = 1.51 V) - `Cr2O7^2−` (E° = 1.38 V) - `Cl2` (E° = 1.40 V) Both `MnO4^−` and `Cr2O7^2−` can oxidize Fe²⁺ to Fe³⁺. **Hint:** Compare the standard reduction potentials to determine which species can act as a stronger oxidizing agent. ### Step 3: Analyze the use of `KMnO4` in acidic medium When `KMnO4` is used in an acidic medium (like HCl), it can react with chloride ions (Cl⁻) instead of oxidizing Fe²⁺. The half-reaction for Cl⁻ is: - `Cl2 + 2e⁻ → 2Cl⁻` (E° = 1.40 V) Since the E° for the reduction of `MnO4^−` is higher than that for Cl⁻, the reaction will favor the oxidation of Cl⁻ to Cl₂ rather than Fe²⁺ to Fe³⁺. **Hint:** Check the E° values to see which reaction is more favorable in the presence of competing species. ### Step 4: Analyze the use of `K2Cr2O7` in acidic medium On the other hand, when `K2Cr2O7` is used in acidic medium, it does not have a competing reaction with Cl⁻ because its E° is lower than that of Cl₂. Therefore, it can effectively oxidize Fe²⁺ to Fe³⁺. **Hint:** Look for the potential reactions that can occur and determine if they are spontaneous based on E° values. ### Step 5: Conclusion The only correct statement regarding the quantitative estimation of aqueous `Fe(NO3)2` is that `KMnO4` cannot be used in the presence of HCl because it will preferentially oxidize Cl⁻ to Cl₂ rather than oxidizing Fe²⁺ to Fe³⁺. In contrast, `K2Cr2O7` can be used effectively for this purpose. **Final Answer:** The only correct statement is that `KMnO4` cannot be used with HCl for the oxidation of `Fe(NO3)2`.
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