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Assertion $(A)$ : Experimental reaction of $CH _{3} Cl$ with aniline and anhydrous $AlCl _{3}$ does not give $o$ and $p-methylaniline$.
Reason $(R)$ : The - $NH _{2}$ group of aniline becomes deactivating because of salt formation with anhydrous $AlCl _{3}$ and hence yields $m$-methyl aniline as the product.
In the light of the above statements, choose the most appropriate answer from the options given below.
$PhC{{O}_{2}}H\xrightarrow[2.\,N{{H}_{3}}]{1.\,PC{{l}_{5}}}A\xrightarrow[2.\,{{H}_{2}}/Ni]{1.\,{{P}_{4}}{{O}_{10}}.heat}B.$ The final product $(B)$ is
$Pt/ M/M^{3+}(0.001 \,mol\, L^{ -1})/Ag^+(0.01\, mol\, L^{-1})/Ag$
The emf of the cell is found to be $0.421\, volt$ at $298\, K$. The standard potential of half reaction $M^{3+} + 3e \to M$ at $298\, K$ will be .............. $\mathrm{volt}$
(Given $E_{A{g^ + }/Ag}^o $ at $298\, K\, = 0.80\, Volt$ )