Question

The standard Electrode Potential values $ \left( {E^\circ } \right) $ of $ A{l^{3 + }}/Al $ , $ A{g^ + }/Ag $ , $ {K^ + }/K $ and $ C{r^{3 + }}/Cr $ are $ - 1.66V,0.80V, - 2.93V, - 0.74V $ respectively. The correct decreasing order of reducing power of the metal is:
(A) $ Ag > Cr > Al > K $
(B) $ K > Al > Cr > Ag $
(C) $ K > Al > Ag > Cr $
(D) $ Al > K > Ag > Cr $

Answer 1

Hint :Standard Electrode potential represents the amount of energy exchanged per mole when the species in the left side converts to the species in the right side of the slash symbol, here in our question it denotes the Standard reduction Potential of ions wherein it represents the energy exchanged per mole when ions undergo reduction.

Complete Step By Step Answer:
The first point to be noted here are the values of $ E^\circ $ . The direction of values i.e. positive or negative represents how favorable is the reduction of the ion provided with. More the positive value is favorable is the reduction of the ion and more unfavorable would be to convert the metal back to the ion.
Here our question asks us to arrange in decreasing order the reducing power of metals which means that the metal should assist other species in reduction thereby itself getting oxidized in the process which means we should consider the standard oxidation potential of metals.
It can be easily obtained by just reversing the nature of standard reduction potential values. Thus, oxidation potential values for each element will be:
 $ Al $ = $ 1.66V $
 $ Ag $ = $ - 0.80V $
 $ K $ = $ 2.93V $
 $ Cr $ = $ 0.74V $
As clearly observed now we know that the metal with the highest reducing power is the one which has the highest positive value of oxidation potential i.e. $ K $ followed by $ Al $ , $ Cr $ and the last being $ Ag $ .
Thus Option (B) $ K > Al > Cr > Ag $ is the correct answer.

Note :
In the present question, just by looking at the positive standard reduction potential of $ Ag $ we knew that it would have the lowest standard oxidation potential among all other provided metals and thus it should come at last in the correct option which it did only in case of option (B) which is the correct answer.