Question

Arrange the following metals in increasing order of their reducing power [Given: $E_{\dfrac{{{K^ + }}}{K}}^0 = --2.93V$ , $E_{Ag}^0 = + 0.80V$ , $E_{\dfrac{{A{l^{ + 3}}}}{{AI}}}^0 = - 1.66V$ . $E_{\dfrac{{A{u^{ + 3}}}}{{Au}}}^0 = + 1.40V$ , $E_{\dfrac{{L{i^ + }}}{{Li}}}^0 = - 3.05V$
A. Li < K < Al < Ag < Au
B. Au < Ag < AI < K < Li
C. K < Al < Au < Ag < Li
D. Al < Ag < Au < Li < K

Answer 1

An oxidizing agent can be explained as a reactant in a redox reaction, that causes other substances to lose an electron, and then accept these electrons for itself. A reducing agent performs the exact opposite operation to this. A reducing agent itself loses its electrons and makes them available for getting accepted by other reactants.

Complete step by step answer:
Before we move forward with the solution of this question, let us first understand some basic concepts.
Reduction potential: Reduction potential can be understood as the tendency of an element or a compound to gain electrons, i.e. undergo reduction. To put it in simpler terms, it is the extent to which an atom or a molecule can be reduced.
Now, the higher the value of the reduction potential of metal lower will be the reducing power and vice-versa.
Therefore, considering the given values of the reduction potential of the metals the order of reducing power is,
Au < Ag < AI < K < Li.

So, the correct answer is, B.

Note:
The oxidizing nature of any species is dependent on the reduction potential of the given species. The higher the reduction potential, the higher would be the oxidizing nature of the species. From the data above, we can say that Fe or iron has the highest reduction potential, and hence is the best oxidizing agent out of the given options.