1. Tardigrade
  2. Question
  3. Chemistry
  4. Electrode potential data are given below. Fe3+ (aq) + e- → Fe2+ (aq); Eo = + 0.77 Al3+ (aq) + 3e- → Al (s); Eo = - 1.66 V Br2 (aq) + 2e- → 2Br- (aq); Eo = +1.08 V Based on the data given above, reducing power of Fe2+, Al and Br- will increase in the order

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Q. Electrode potential data are given below.
$Fe^{3+} (aq) + e^{-} \to Fe^{2+} (aq); \,\,\,E^{o} = + 0.77$
$Al^{3+} (aq) + 3e^{-} \to Al (s); E^{o} \,\,\,= - 1.66 V$
$Br_{2} (aq) + 2e^{-} \to 2Br^{-} (aq); \,\,\, E^{o} = +1.08 V$
Based on the data given above, reducing power of $Fe^{2+}, Al$ and $Br^{-}$ will increase in the order

Electrochemistry

Solution:

The lowest the standard reduction potential, the highest is the reducing power.