Identify the correct statement(s):

• Correct answer
  A

  The oxidation number of $\mathrm{Cr}$ in $\mathrm{CrO}_{5}$ is +6
• Correct answer
  B

  $\Delta H>\Delta U$ for the reaction $N_{2} O_{4}(g) \rightarrow 2 N O_{2}(g)$ provided both gases behave ideally.
• C

  $\mathrm{pH}$ of $0.1 \mathrm{N} \mathrm{H}_{2} \mathrm{SO}_{4}$ is less than that of $0.1 \mathrm{N} \mathrm{HC}$ at $25^{\circ} \mathrm{C}$
• D

  $\left(\frac{R T}{F}\right)=0.0591$ volt at $25^{\circ} \mathrm{C}$

Solution

In $\mathrm{CrO}_{5}$ oxidation no. of $\mathrm{Cr}$ is +6. (b) For the reaction (in ideal case) $$ \begin{array}{r} \mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\ \Delta H=\Delta U+p \cdot \Delta V \end{array} $$ $\because \quad \Delta H=\Delta U+\Delta n_{g} R T$ $\left(\because p \Delta V=\Delta n_{g} R T\right)$ Where, $\Delta n_{g}=$ (gaseous moles of product)-(gaseous moles of reactant) $$ \Delta n_{g}=2-1=1 $$ Thus, $\Delta H>\Delta U$ (c) ${p H}$ of $0.1 \mathrm{N} \mathrm{H}_{2} \mathrm{SO}_{4}$ is less than of $0.1 \mathrm{HCl}$ at $25^{\circ}$ is a wrong statement. $\mathrm{pH}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ $$ \because \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left[\mathrm{l} 0^{-2}\right] $$ $\because \quad(N=C \times Z)$ and $\left(Z=2\right.$ for $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$ $\mathrm{pH}=2$ and $\mathrm{pH}$ of $\mathrm{HCl}$ $$ \mathrm{pH}=-\log \left[\mathrm{l} 0^{-1}\right]=1 $$ Thus, $\mathrm{pH}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}>\mathrm{pH}$ of $\mathrm{HCl}$. (d) $\frac{R T}{F}=0.0591$ at $25^{\circ} \mathrm{C}$ is a wrong statement. Where, $R=$ Gas constant $=8.314$ $T=$ Temperature $=273+25=298 \mathrm{K}$ $F=$ Charge over one mole of electrons $=96500 \mathrm{C}$ Thus, $\frac{R T}{F}=\frac{8.314 \times 298}{96500}=0.02567$ The correct value is $\frac{2.303 R T}{F}=0.0591$ Hence, only option and (b) are correct.

Hence, the correct answer is options: 'A', 'B'.