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$Fe ^{2+} \rightarrow Fe ^{3+} + e ^{-} \quad E _{ Fe ^{3+} / Fe ^{2+}}=0.77 \,V$
$2 I ^{-} \rightarrow I _{2}+2 e ^{-} \quad E _{ I _{2} / I ^{-}}^{0}=0.54 \,V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \,V 298$ $K$. The value of $x$ is .... (Nearest Integer)
$\mathrm{Pt}(s) \mid \mathrm{H}_2(g, 1 \text { bar })\left|\mathrm{H}^{+}(a q, 1 \mathrm{M}) \| \mathrm{M}^{4+}(a q), \mathrm{M}^{2+}(a q)\right| \mathrm{Pt}(s)$
$E_{\text {cell }}=0.092 \mathrm{~V} \text { when } \frac{\left[\mathrm{M}^{2+}(a q)\right]}{\left[\mathrm{M}^{4+}(a q)\right]}=10^x$
Given : $ E_{\mathrm{M}^4 / \mathrm{M}^{2+}}^0=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}$
The value of $x$ is
Statement $I$: The boiling point of hydrides of Group $16$ elements follow the order
$\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}$.
Statement $II$: On the basis of molecular mass, $\mathrm{H}_2 \mathrm{O}$ is expected to have lower boiling point than the othe members of the group but due to the presence of extensive $\mathrm{H}$-bonding in $\mathrm{H}_2 \mathrm{O}$, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
$\left[\Lambda_{\mathrm{m}}=\right.$ molar conductivity
$\Lambda_{\mathrm{m}}^{\circ}=$ limiting molar conductivity
$\mathrm{c}=$ molar concentration
$\mathrm{K}_{\mathrm{a}}=$ dissociation constant of $\mathrm{HX}$ ]
$(CH_3)_2CHCH_2N(CH_2CH_3)_2 \xrightarrow{C{{H}_{3}}I}\xrightarrow[{{H}_{2}}O]{A{{g}_{2}}O}\xrightarrow{heat}$ products