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Explain why Cr^(2+) is strongly reducing while Mn^(3+) is strongly oxidizing. |
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Answer» Solution :`Cr^(2+)` is STRONG reducing while MN3+ is strongly oxidising, `E^(0)_(Cr^(3+)//Cr^(2+))` is-0.41V `Cr^(2+)+ 2e^(-) to Cr E^(0)= -091 V` If the standard electrode potential `E^(0)` of a metal is large and negative, the metal is a powerful reducing agent because it LOSES electrons easily. `Mn to Mn^(3+)+ 3e^(-)Mn^(3+) [Ar]3d^(4)` `Mn^(3+)+ e^(-) to Mn^(2+)E^(0) = +1.51V` If the standard electrode potential Eº of a metal is large and positive, the metal is a powerful oxidising agent because it gains electrons easily, |
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