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Standard reduction potentails of the hal...

Standard reduction potentails of the half reactions are given below:
`F_(2)(g)+2e^(-) rarr 2F^(-)(aq.),, E^(ɵ)= +2.87`
`Cl_(2)(g)+2e^(-) rarr 2Cl^(-)(aq.),, E^(ɵ)= +1.36 V`
`Br_(2)(g)+2e^(-) rarr 2Br^(-)(aq.),, E^(ɵ)= +1.09 V`
`I_(2)(s)+2e^(-) rarr 2l^(-)(aq.),, E^(ɵ)= +0.54 V`
The strongest oxidizing and reducing agents respectively are:

A

`F_(2) and I^(-)`

B

`Br_(2) and Cl^(-)`

C

`Cl_(2) and Br^(-)`

D

`Cl_(2) and I_(2)`

Text Solution

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The correct Answer is:
To determine the strongest oxidizing and reducing agents from the given standard reduction potentials, we can follow these steps: ### Step 1: Understand the concept of standard reduction potentials Standard reduction potentials (E°) indicate the tendency of a species to gain electrons and be reduced. A higher E° value means a stronger oxidizing agent, while a lower E° value indicates a stronger reducing agent. ### Step 2: List the given half-reactions and their E° values - \( F_2(g) + 2e^- \rightarrow 2F^-(aq), \quad E° = +2.87 \, V \) - \( Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq), \quad E° = +1.36 \, V \) - \( Br_2(g) + 2e^- \rightarrow 2Br^-(aq), \quad E° = +1.09 \, V \) - \( I_2(s) + 2e^- \rightarrow 2I^-(aq), \quad E° = +0.54 \, V \) ### Step 3: Identify the strongest oxidizing agent The strongest oxidizing agent corresponds to the half-reaction with the highest E° value. From the list: - The highest E° value is +2.87 V for the reaction involving \( F_2 \). Thus, \( F_2 \) is the strongest oxidizing agent. ### Step 4: Identify the strongest reducing agent The strongest reducing agent corresponds to the half-reaction with the lowest E° value. From the list: - The lowest E° value is +0.54 V for the reaction involving \( I_2 \). Thus, \( I^- \) is the strongest reducing agent. ### Conclusion The strongest oxidizing agent is \( F_2 \) and the strongest reducing agent is \( I^- \). ### Final Answer - Strongest oxidizing agent: \( F_2 \) - Strongest reducing agent: \( I^- \) ---
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Knowledge Check

  • Based on the data given below, the correcy order of reducing power is: Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V

    A
    `Br^(-) lt Fe^(2+) lt Al`
    B
    `Fe^(2+) lt Al lt Br^-`
    C
    `Al lt Br^(-) lt Fe^(2+)`
    D
    `Al lt Fe^(2+) lt Br^(-)`

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