Question

Arrange the following metals in increasing order of their reducing power.
[Given : $E^{\circ}_{K^{+}/K} = -2.93\, V$, $E^{\circ} _{Ag^{+}/Ag }= + 0.80 \,V$,
$E^{\circ} _{Al^{3+}/Al}= -1-66 \,V$, $E^{\circ} _{Au^{3+}/Au} = + 1.40 \,V$,
$E^{\circ}_{Li^{+}/Li }=-3.05 \,V$]

• 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)

Lower the electrode potential, better is the reducing power. The electrode potential increases in the order of $Li^+/Li (-3.05\, V)$, $K^+/K (-2.93\, V)$, $Al^{3+}/Al$, $(-1.66 \,V)$, $Ag^+/Ag (+0.80\, V)$ and $Au^{3+}/Au (+1.40\, V)$. Hence, reducing power of metals will be $Au < Ag < Al < K < Li$.