The standard electrode potential ($E^{\circ }$) values of $\mathrm{Al}{}^{3+}/\mathrm{Al}$, $\mathrm{Ag}{}^{+}/\mathrm{Ag}$, $\mathrm{K}{}^{+}/\mathrm{K}$ and $\mathrm{Cr}{}^{3+}/\mathrm{Cr}$ are $-1.66$ V, $0.80$ V, $-2.93$ V and $-0.74$ V, respectively. The correct decreasing order of reducing power of the metals is

Solution & Explanation

\textbf{Answer:} (b) $\mathrm{K}>\mathrm{Al}>\mathrm{Cr}>\mathrm{Ag}$. \textbf{Solution:} The reducing power of a metal is inversely related to its standard reduction potential (SRP): the more negative the SRP, the stronger the reducing agent. Ordering the metals by SRP: $\mathrm{K}$ $(-2.93)<\mathrm{Al}$ $(-1.66)<\mathrm{Cr}$ $(-0.74)<\mathrm{Ag}$ $(+0.80)$. Therefore the reducing power decreases as $\mathrm{K}>\mathrm{Al}>\mathrm{Cr}>\mathrm{Ag}$.