Electrode potential data are given below Fe3+(aq)++e-Fe2+(aq) E° =+0.77V Al3+(aq)+3e- xrightarrow[]Al(s) E° =-1.66 V Br2(aq)+2e- xrightarrow[]2Br-(aq) E° =+1.08 V Based on the data given above, reducing power of Fe2+,Al and Br-, will increase, in the order

Question

Electrode potential data are given below Fe3+(aq)++e-Fe2+(aq) E° =+0.77V Al3+(aq)+3e- xrightarrow[]Al(s) E° =-1.66 V Br2(aq)+2e- xrightarrow[]2Br-(aq) E° =+1.08 V Based on the data given above, reducing power of Fe2+,Al and Br-, will increase, in the order (A)  Br-<Fe2+<Al  (B)  Fe2+<Al<Br-  (C)  Al<Br-<Fe2+  (D)  Al<Fe2+<Br-

Tardigrade Admin · Accepted Answer

Reducing power ie, the tendency to lose electrons increases as the reduction potential decreases so the increasing order of reducing power of the given ions is Br-~<Fe2+<Al .

$B r^{-} < F e^{2 +} < A l$

$F e^{2 +} < A l < B r^{-}$

$A l < B r^{-} < F e^{2 +}$

$A l < F e^{2 +} < B r^{-}$

Reducing power ie, the tendency to lose electrons increases as the reduction potential decreases so the increasing order of reducing power of the given ions is $B r^{-} < F e^{2 +} < A l$ .

Q. Electrode potential data are given below Fe3+(aq)++e−Fe2+(aq) E∘=+0.77V Al3+(aq)+3e−​Al(s) E∘=−1.66V Br2​(aq)+2e−​2Br−(aq) E∘=+1.08V Based on the data given above, reducing power of Fe2+,Al and Br−, will increase, in the order

Br−<Fe2+<Al

Fe2+<Al<Br−

Al<Br−<Fe2+

Al<Fe2+<Br−

Reducing power ie, the tendency to lose electrons increases as the reduction potential decreases so the increasing order of reducing power of the given ions is Br− <Fe2+<Al .

$B r^{-} < F e^{2 +} < A l$

$F e^{2 +} < A l < B r^{-}$

$A l < B r^{-} < F e^{2 +}$

$A l < F e^{2 +} < B r^{-}$

Reducing power ie, the tendency to lose electrons increases as the reduction potential decreases so the increasing order of reducing power of the given ions is $B r^{-} < F e^{2 +} < A l$ .