Standard reduction potentials of the half reactions are given below [AIPMT (M) 2012] |
\[{{F}_{2}}(g)=2{{e}^{-}}\xrightarrow[{}]{{}}2{{F}^{-}}(aq);\]\[{{E}^{o}}=+\,2.85\,V\]\[C{{l}_{2}}(g)+2{{e}^{-}}\xrightarrow[{}]{{}}2C{{l}^{-}}(aq);\]\[{{E}^{o}}=+\,1.36\,V\] |
\[B{{r}_{2}}(l)+2{{e}^{-}}\xrightarrow[{}]{{}}2B{{r}^{-}}(aq);\]\[{{E}^{o}}=+1.06\,V\] |
\[{{I}_{2}}(s)+2{{e}^{-}}\xrightarrow[{}]{{}}2{{I}^{-}}(aq);\]\[{{E}^{o}}=+0.53\,V\] |
The strongest oxidsing and reducing agents respectively are |
A) \[{{F}_{2}}\] and \[{{I}^{-}}\]
B) \[B{{r}_{2}}\] and \[C{{l}^{-}}\]
C) \[C{{l}_{2}}\] and \[B{{r}^{-}}\]
D) \[C{{l}_{2}}\] and \[{{I}_{2}}\]
Correct Answer: A
Solution :
[a] Higher the value of standard reduction potential stronger will be the oxidising agent. Therefore F2 will act as strongest oxidizing agent. |
Similarly lower the value of standard reduction potential stronger will be the reducing agent. |
Therefore \[{{\text{I}}^{-}}\]will act as strongest reducing agent. |
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