Given below are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be `K^(+)//K`

Question

Given below are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be `K^(+)//K`

Given below are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be `K^(+)//K`=-2.93 V , `Ag^+` /Ag=0.80 V, `Mg^(2+)`/Mg=-2.37 V, `Cr^(3+)`/Cr =-0.74 V
A. K lt Mg lt Cr lt Ag
B. Ag lt Cr lt Mg lt K
C. Mg lt K lt Cr lt Ag
D. Cr lt Ag lt Mg lt K

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
Higher the oxidation potential, more easily it is oxidised and hence greater is the reducing power. Hence, increasing order of reducing power is Ag lt Cr lt Mg lt K

4 views Answered 2023-02-05 15:29