Question

Given the standard electrode potentials,
${K^ + }/K = - 2.93V,A{g^ + }/Ag = 0.80 V$
$H{g^{2 + }}/Hg = 0.79 V$
$M{g^{2 + }}/Mg = - 2.37V,Cr^{3 + }/Cr = - 0.74 V$
Arrange these metals in their increasing order of reducing power

Answer 1

Hint: The tendency to gain electron means also to get reduced. This tendency is called the reduction potential. Similarly, tendency to lose an electron means the tendency to get oxidised. This tendency is called the oxidation potential.

Complete step by step answer:
To compare the relative reducing powers. As by convention, negative sign is used to represent oxidation potential, this implies that a large negative value of oxidation potential is more easily the substance (element or ion) oxidised or in other words, stronger reducing agent it is. In given question the oxidation potential
${K^ + }/K = - 2.93V, A{g^ + }/Ag = 0.80 V$
$H{g^{2 + }}/Hg = 0.79 V$
$M{g^{2 + }}/Mg = - 2.37V, Cr^{3+}/Cr = - 0.74 V$
From this question it is clear that the oxidation potential of ${K^ + }/K = - 2.93 V$ which is largest negative, so it is a most powerful reducing agent and $A{g^ + }/Ag = 0.80 V$ which is largest positive means lowest oxidation potential. Hence, it is the weakest reducing agent in the above question. Hence, the decreasing order of reducing power given below
$K > Mg > Cr > Hg > Ag$
Hence the decreasing order of reducing power is:
$K^+ > Mg^{2+} > Cr^{3+} > Ag^{+1}$

Note: A standard reduction potential is measured using a galvanic cell which contains a SHE on one side and unknown chemical half-cell on the other side. The amount of charge that passes between the cells is measured using a voltmeter.
The reduction potential for the reverse of the oxidation half – reaction and reverse the sign to obtain the oxidation potential. For the oxidation half – reaction, ${E^0}$ oxidation $ = - {E^0}$ reduction. Add the potentials of the half – cells to get the overall standard cell potential.