Question

Arrange the following metals in increasing order of their reducing power.

Given:

${E^{o}}_{K^{+}/K} = - 2.93V,{E^{o}}_{Ag^{+}/Ag} = + 0.80V,{E^{o}}_{Al^{3 +}/Al}$

$= - 1.66V,{E^{o}}_{Au^{3 +}/Au} = + 1.40V,{E^{o}}_{Li^{+}/Li} = - 3.05V$

• A. Li < K < Al < Ag < Au
• B. Au < Ag < Al < K < Li
• C. K < Al < Au < Ag < Li
• D. Al < Ag < Au < Li < K

Answer 1

Answer: (B)

Au < Ag < Al < K < Li

Answer 2

: Lower the electrode potential, better is the reducing power. The electode potential increases in the order of

}{Al^{3 +}/Al,( - 1.66V),}$$ $Ag^{+}/Ag(\_ 0.80V)$ and $Au^{3 +}/Au( + 1.40V).$ Hence, reducing power of metals will be Au, \< Ag \< Al \< K \< Li