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)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,