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Electrode potential data are given below...

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

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
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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 :

Electrode potential data are given below {:(Fe^(3+)(aq)+e^(-)rarrFe^(2+)(aq),E^(@)=+0.77V),(Al^(3+)(aq)+3e^(-)rarrAl(s),E^(@)=-1.66V),(Br_(2)(aq)+2e^(-)rarrBr^(-)(aq),E^(@)=+1.05V):} Based on the given data which statements is/are true?

Given : Fe^(2+)(aq)+2e^(-) rarr Fe(s), " "E^(c-)=-0.44V Al^(3)+3e^(-) rarr Al(s)," "E^(c-)=-1.66V Br^(2+)+2e^(-)rarr 2Br^(c-)(aq)" "E^(c-)=-1.08V The decreasing order of reducing power is ……………………………….. .

Al^(3+)(aq)+3e^(-)rarrAl(s) E^(@)=-1.66V Mn^(2+)(aq)+2e^(-)rarrMn(s) E^(@)=-1.18V What is the standard potential of a volaic cell produced by using these two half-reactions?

Al^(3+)(aq)+3e^(-)rarrAl(s) E^(@)=-1.66V Mn^(2+)(aq)+2e^(-)rarrMn(s) E^(@)=-1.18V What process occurs at the anode of a volatic cell utilizing these two half-reactions?

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