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Standard reduction potentials of the half reactions are given below:${{F}_{2}}_{(g)}+2{{e}^{-}}\to 2{{F}^{-}}_{(aq)}$; ${{E}_{o}}=+2.85V$$C{{l}_{2}}_{(g)}+2{{e}^{-}}\to 2C{{l}^{-}}_{(aq)}; {{E}_{o}}=+1.36V$$B{{r}_{2}}_{(g)}+2{{e}^{-}}\to 2B{{r}^{-}}_{(aq)}$; ${{E}_{o}}=+1.06V$${{I}_{2}}_{(g)}+2{{e}^{-}}\to 2{{I}^{-}}_{(aq)}$; ${{E}_{o}}=+0.53V$The strongest oxidizing and reducing agents respectively are:A. $B{{r}_{2}}$ and $C{{l}^{-}}$B. $C{{l}_{2}}$ and $B{{r}^{-}}$C. $C{{l}_{2}}$ and ${{I}_{2}}$D. ${{F}_{2}}$ and ${{I}^{-}}$

Last updated date: 13th Jun 2024
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Complete Solution :The standard reduction potential can be referred to as a standard hydrogen reference electrode which immediately gives a potential of 0.00 volts. The values below in parentheses are standard reduction potentials for half-reactions which are measured at ${25 ^o}$C, 1 atmosphere and with a pH of 7 in aqueous solution. As we know that reduction potential is used to measure the intrinsic tendency for a species to undergo reduction as compared to standard reduction potential for two processes and it can be useful for determining how a reaction will proceed. Each half-cell is associated with a potential difference whose value depends on the nature of the particular electrode reaction and also on the concentrations of the dissolved species. The sign of this potential difference depends on the direction in which the electrode reaction proceeds.
- Higher the value of standard electrode potential stronger will be the oxidizing agent so we can say that ${{F}_{2}}$ is the stronger oxidizing agent. Similarly lower the value of standard electrode potential stronger will be the reducing agent so ${{I}^{-}}$ will be the strongest reducing agent out of the other mentioned in the question. Thus, option D is correct.