Electrochemistry Question 5
Question 5 - 29 January - Shift 1
The standard electrode potential $(M^{3+} / M^{2+})$ for $V$, $Cr$, Mn & Co are $-0.26 V,-0.41 V,+1.57 V$ and $+1.97 V$, respectively. The metal ions which can liberate $H_2$ from a dilute acid are
(1) $V^{2+}$ and $Mn^{2+}$
(2) $Cr^{2+}$ and $CO^{2+}$
(3) $V^{2+}$ and $Cr^{2+}$
(4) $Mn^{2+}$ and $Co^{2+}$
Show Answer
Answer: (3)
Solution:
Metal cation with $(-)$ value of reduction potential $(M^{+3} / M^{+2})$ or with $(+)$ value of oxidation potential $(M^{+2} / M^{+3})$ will liberate $H_2$
Therefore they will reduce $H^{+}$
i. $eV^{+2}$ and $Cr^{+2}$
