Electrode potentials (E^o) are given below Cu^+ /Cu = + 0.52 , V, Fe^ 3+ /Fe^ 2+ = +0.7 7 , V, 1 2 I_2 ( s )/ I^ - , = + 0.54 ,V, Ag^+ /Ag = + 0.88 ,V. Based on the above potentials, strongest oxidizing agent will be

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V}$

$\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \mathrm{E}^{\circ}=-0.04 \mathrm{~V}$

$\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \mathrm{E}^{\circ}=-0.25 \mathrm{~V}$

$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \mathrm{E}^{\circ}=0.80 \mathrm{~V}$

$\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \mathrm{E}^{\circ}=1.40 \mathrm{~V}$

Consider the given electrochemical reactions, The number of metal$(s)$ which will be oxidized be $\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}$, in aqueous solution is. . . . . .

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If two substances $A$ and $B$ have $P_A^0:\,P_B^0 = 1\,:\,2$ and have mole fraction in solution $1 : 2 $ then mole fraction of $A$ in vapours

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Vitamin ${B_{12}}$ contains metal

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The end product B in the sequence of reactions

$R - X\xrightarrow{{KCN}}A\mathop {\xrightarrow{{NaOH}}}\limits_{{H_2}O} $ $B$ is

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The molar freezing point constant for water is ${1.86\,^o}C\,mol{e^{ - 1}}$. If $ 342\,gm$ of canesugar $({C_{12}}{H_{22}}{O_{11}})$ are dissolved in $1000\,gm$ of water, the solution will freeze at ............ $^oC$.

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The reaction($s$) leading to the formation of $1,3,5-$trimethylbenzene is (are)

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