Question

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

• 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)

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