The given problem involves determining which metals will be oxidized by the dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\). The dichromate ion reduction potential is \(E^\circ = 1.33 \, \text{V}\). To determine if a metal will be oxidized, compare its standard reduction potential to that of the dichromate ion. A metal with a lower (more negative) reduction potential will be oxidized by the dichromate ion.
Let's analyze each half-reaction:
Since \(-0.04 \, \text{V} < 1.33 \, \text{V}\), \(\text{Fe}\) can be oxidized.
Since \(-0.25 \, \text{V} < 1.33 \, \text{V}\), \(\text{Ni}\) can be oxidized.
Since \(0.80 \, \text{V} < 1.33 \, \text{V}\), \(\text{Ag}\) can be oxidized.
Since \(1.40 \, \text{V} > 1.33 \, \text{V}\), \(\text{Au}\) cannot be oxidized.
Thus, three metals—\(\text{Fe}\), \(\text{Ni}\), and \(\text{Ag}\)—will be oxidized by \(\text{Cr}_2\text{O}_7^{2-}\). The result, 3, confirms that the solution falls within the provided range (3,3).
Metals with lower standard reduction potentials (Eo) compared to Cr2O72− (Eo = 1.33 V) will be oxidized. These are:
Thus, the number of metals oxidized is 3.
Final Answer: (3)
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.