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)

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming $M^+$ ($M \to M^+ + e^-$). The cation $M^+$ is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]


A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 \(\Omega\) then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _____ N.