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Given    Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq);E^{\circ}=+0.77\, V

Al^{3+}(aq)+3e^{-}\rightarrow Al(s);E^{\circ}=-1.66\, V

Br_{2}(aq)+2e^{-}\rightarrow 2Br^{-};E^{\circ}=+1.09V

Considering the electrode potentials, which of the following represents the correct order of reducing power ?

Option: 1

Fe^{2+}< Al< Br^{-}


Option: 2

Br^{-}< Fe^{2+}< Al


Option: 3

Al< Br^{-}< Fe^{2+}


Option: 4

Al< Fe^{2+}< Br^{-}    


Answers (1)

best_answer

As learned in concept

Standard reduction potential -

It measure the tendency of a chemical species to acquire electrons and there by be reduced. It is measured in volts or mv.

-

 

 Fe^{3+}(aq) +e^-\rightarrow Fe^{3+}(aq), E^0=+0.77V

 

Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) +e^- , E^0 = -0.77V ...........equ(1)

also AI(s)\rightarrow AI^{3+}(aq) +3e^-, E^0=-1.66V .................equ\:(2)

and finally Br_2 +2e^-\rightarrow 2Br^-, E^0=+1.09V

= 2Br^- \rightarrow Br_2(aq) + 2e^-, E^0= - 1.09V ..............equ (3)

In reaction 1,2 and 3 the higher the value of E^0 the greater the reducing power

\therefore A_1>F_e^{2+}>B_r^-

as 1.66V> -0.77V> > -1.09V

Answer is 2

Posted by

seema garhwal

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