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, Mg2+/Mg=-2.37 V, Cr3+/Cr=-0.74 V.

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=-2.93 V, Ag+/Ag=0.80 V, Mg2+/Mg=-2.37 V, Cr3+/Cr=-0.74 V. (A) K < Mg < Cr < Ag (B) Ag < Cr < Mg < K (C) Mg < K < Cr < Ag (D) Cr < Ag < Mg < K

Tardigrade Admin · Accepted Answer

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

Q. Given below are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be
$K^{+} / K = - 2.93 V$ , $A g^{+} / A g = 0.80 V$ ,
$M g^{2 +} / M g = - 2.37 V$ , $C r^{3 +} / C r = - 0.74 V$ .

$K < M g < C r < A g$

$A g < C r < M g < K$

$M g < K < C r < A g$

$C r < A g < M g < K$

Higher the oxidation potential, more easily it is oxidised and hence greater is the reducing power. Hence, increasing order of reducing power is
$A g < C r < M g < K$

Q. 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=−2.93V, Ag+/Ag=0.80V, Mg2+/Mg=−2.37V, Cr3+/Cr=−0.74V.

K<Mg<Cr<Ag

Ag<Cr<Mg<K

Mg<K<Cr<Ag

Cr<Ag<Mg<K