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

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^(-)`

Text Solution

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The correct Answer is:
A
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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

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

Standard reduction potentials of the half reactions are given below {:(F_(2(g))+2e^(-) rarr 2F^(-)""_((aq)),,E^(@)=+2.85V), (Cl_(2(g))+2e^(-) rarr 2Cl^(-)""_((aq)),, E^(@)=+1.36V), (Br_(2(l))+2e^(-) rarr 2Br^(-)""_((aq)),,E^(@)=+1.06V), (l_(2(s))+2e^(-) rarr 2l^(-)""_((aq)),, E^(@)=+0.53V):} The strongest oxidising and reducing agents respectively are

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:

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