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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^(-)` |
Answer» Correct Answer - A (a) Electrode potentials in this case are standard oxidation potentials. The standard reduction potentials have opposite signs i.e. `Fe^(2+)(aq) to Fe^(3+)(aq)+e^(-),E^(@)=-0.77" V "` `Al(s) to Al^(3+)(aq)+3e^(-),E^(@)=+1.66" V "` `2Br^(-)(aq) to Br_(2)(aq)+2e^(-),E^(@)=-1.08" V "` Reducing power is in the order : `Br^(-) lt Fe^(2+) lt Al`. |
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