Question

The standard reduction potentials at 298 K for the following half cells are given below:
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}, \, E^\circ = 1.33 \, \text{V}$
$\text{Fe}^{3+} (\text{aq}) + \text{3e}^- \rightarrow \text{Fe}, \, E^\circ = -0.04 \, \text{V}$
$\text{Ni}^{2+} (\text{aq}) + 2\text{e}^- \rightarrow \text{Ni}, \, E^\circ = -0.25 \, \text{V}$
$\text{Ag}^+ (\text{aq}) + \text{e}^- \rightarrow \text{Ag}, \, E^\circ = 0.80 \, \text{V}$
$\text{Au}^{3+} (\text{aq}) + 3\text{e}^- \rightarrow \text{Au}, \, E^\circ = 1.40 \, \text{V}$
Consider the given electrochemical reactions, the number of metal(s) which will be oxidized by $\text{Cr}_2\text{O}_7^{2-}$ in aqueous solution is ____.

Answer 1

Metals with lower standard reduction potentials (E o) compared to Cr2O72− (E o = 1.33 V) will be oxidized. These are:

• Fe (E o = −0.04 V),
• Ni (E o = −0.25 V),
• Ag (E o = 0.80 V).

Thus, the number of metals oxidized is 3.

Final Answer: (3)