Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be $1 / 30$ of the molar conductivity of another weak acid HZ with concentration of 0.02 M . If $\lambda^{\circ} \mathrm{Q}^{-}$happened to be equal with $\lambda^{\circ} \mathrm{Z}^{-}$, then the difference of the $\mathrm{pK}_{\mathrm{a}}$ values of the two weak acids $\left(\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ})-\mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})\right)$ is $\_\_\_\_$ (Nearest integer).

[Given: degree of dissociation $(\alpha) \ll 1$ for both weak acids, $\lambda^{\circ}$ : limiting molar conductivity of ions]

Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL . The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL . (Nearest integer)

[Given:

$$ \begin{aligned} & \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{~s}) \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=+0.34 \mathrm{~V} \\ & \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=+1.23 \mathrm{~V} \end{aligned} $$

Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of $\mathrm{O}_2=32 \mathrm{~g} \mathrm{~mol}^{-1}$

Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$

Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$ ]

Consider the following electrochemical cell :

$$ \mathrm{Pt}\left|\mathrm{O}_2(\mathrm{~g})(1 \mathrm{bar})\right| \mathrm{HCl}(\mathrm{aq}) \| \mathrm{M}^{2+}(\mathrm{aq}, 1.0 \mathrm{M}) \mid \mathrm{M}(\mathrm{~s}) $$

The pH above which, oxygen gas would start to evolve at anode is $\_\_\_\_$ (nearest integer).

$$ \left.\left[\begin{array}{ll} \text { Given : } & \mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{o}}=0.994 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\mathrm{o}}=1.23 \mathrm{~V} \end{array}\right\} \text { standard reduction potential } \\ \text {and} \frac{\mathrm{RT}}{\mathrm{F}}(2.303)=0.059 \mathrm{~V} \text {at the given condition}\right] $$

Consider the following electrochemical cell at 298 K

$\mathrm{Pt}\left|\mathrm{HSnO}_2^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_6{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_2 \mathrm{O}_3(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$.

If the reaction quotient at a given time is $10^6$, then the cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is

$\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer).

Given the standard half-cell reduction potential as

$$ \mathrm{E}_{\mathrm{Bi}_2 \mathrm{O}_3 / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_6^{2-} / \mathrm{HSnO}_2^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V} $$