Question

Which of the following(s) can act as strong reducing agents?

• A. $T{i}^{2+}$
• B. $V^{2+}$
• C. $C{r}^{2+}$
• D. $C{o}^{2+}$

Answer 1

Answer: (A), (B), (C)

$C{r}^{2+}$

The value of $E_{M^{3+}/M^{2+}}^0$ of 3d elements increases across the period because the third ionisation energy increase increase across the period.

Reducing agents will reduce other compounds and itself get oxidised. Thus, high negative value of $E_{M^{3+}/M^{2+}}^0$ indicates its less ability to get reduced from $M^{3+}$ to $M^{2+}$.
Also, high negative value of $E_{M^{3+}/M^{2+}}^0$ indicates its ability to get oxidised from $M^{2+}$ to $M^{3+}$.
Hence, ions with high negative value of $E_{M^{3+}/M^{2+}}^0$ will act as strong reducing agent.

In 3d-series, $Ti,V$ and $Cr$ has high negative $E_{M^{3+}/M^{2+}}^0$.
Hence, $T{i}^{2+},V^{2+}$ and $C{r}^{2+}$ act as strong reducing agents.

Cobalt has high $E_{M^{3+}/M^{2+}}^0$ value due to higher third ionisation value.

Hence, $C{o}^{3+}$ will act as strong oxidising agents.