Question

Metal	Electrode potential (V)
$N{a}^{+}$	- 2.71
$M{g}^{2+}$	-2.37
$A{l}^{3+}$	-1.66
$P{b}^{2+}$	-0.13

(i) Which of the above mentioned metal ions will be oxidized first ?
(ii) Which of the above mentioned metal ions will be reduced first?

• A. (i)$N{a}^{+}$(ii)$P{b}^{2+}$
• B. (i)$N{a}^{+}$(ii)$A{l}^{3+}$
• C. (i)$P{b}^{2+}$(ii)$M{g}^{2+}$
• D. (i)$P{b}^{2+}$(ii)$N{a}^{+}$

Answer 1

The correct option is A (i)$N{a}^{+}$

(ii)$P{b}^{2+}$

Metals which are at the top of the electrochemical series have high value of negative electrode potential and they are good reducing agents. They give away electrons easily. And thus they get oxidized easily too. Among the above mentioned metal ions, $N{a}^{+}$ has the highest negative electrode potential. Thus it loses electron most easily and gets oxidized first.
Metals which are towards the bottom of the electrochemical series have positive value of electrode potential and they are good oxidizing agents. They accept electrons very easily. And thus they get reduced easily too. Among the above mentioned metal ions, $P{b}^{2+}$ has the lowest negative electrode potential. Thus it accepts electron most easily and gets reduced first.