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Given the standard electrode potentials, K^(+)//K = -2.93 V, Ag^(+)//Ag = 0.80 V, Hg^(2+)//Hg = 0.79 V Mg^(2+)//Mg = -2.37 V, Cr^(3+)//Cr = -0.74 V Arrange these metals in their increasing order of reducing power. |
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Answer» Solution :More negative the reduction potential, more easily it is oxidised and HENCE greater is the reducing power. THUS, INCREASING order of reducing power will be : `Ag lt Hg lt Cr lt Mg lt K` |
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