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

Options

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 `

Answer

B

Explanation

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