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Based on the data given below, the corre...

Based on the data given below, the correct 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^(-)`

Text Solution

Verified by Experts

The correct Answer is:
A
Lower the reduction potential, more is the reducing power .
`Br^-` lt `Fe^(2+)` lt Al
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Standard electrode potential data is given below : Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V Based on the data given above, reducing power of Fe^(2+) Al and Br^(-) will increase in the order :

Based on data given below for the electrode potential , reducing power of Fe^(2+) , Al and Br^(-) will increase in the order Br_(2)(aq)+2e^(-)rarr2Br^(-)(aq),E^@=+1.08V Al^(3+)(aq)+3e^(-)rarrAl(s),E^@=-1.66V Fe^(3+)(aq)+e^(-2)rarrFe^(+2) (aq) , E^(@) =+0.77V

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Standrd reduction potentials of the half reactions are given below : F_2 (g) +re^- rarr 2 F^(-) (aq) E^@ = + 2.85 V Cl_2 (g) +2e^- rarr 2 Cl^-(aq) , E^2 = + 1.36 V Br _2 (i) + 2 e^- rarr 2Br (aq) , E^2 = + 1. 06 V I_2 (s) + 2 e^- rarr 2I^(-) (aq) , E^2 = + . 53 V . The strongest oxidizing and reducing agents respectively

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