At $$291 \mathrm{~K}$$, saturated solution of $$\mathrm{BaSO}_4$$ was found to have a specific conductivity of $$3.648 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$$ and that of water being used is $$1.25 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$$. If the ionic conductances of $$\mathrm{Ba}^{2+}$$ and $$\mathrm{SO}_4^{2-}$$ are 110 and $$136.6 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$ respectively. The solubility of $$\mathrm{BaSO}_4$$ at $$291 \mathrm{~K}$$ will be [Atomic masses of $$\mathrm{Ba}=137, \mathrm{~S}=32, \mathrm{O}=16]$$

Find the emf of the following cell reaction. Given, $$E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\Upsilon}=-0.72 \mathrm{~V}$$ and $$E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\Upsilon}= -0.42 \mathrm{~V}$$ at $$25^{\circ} \mathrm{C}$$ is $$\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M})\right| \mid \mathrm{Fe}^{2+} (0.1 \mathrm{M}) \mid \mathrm{Fe}$$

For $$\mathrm{C{r_2}O_7^{2 - } + 14{H^ + } + 6{e^ - }\buildrel {Yields} \over \longrightarrow 2C{r^{3 + }} + 7{H_2}O,{E^\Upsilon } = 1.33}$$ V at $$[C{r_2}O_7^{2 - }] = 4.5$$ millimole, $$[C{r^{3 + }}] = 1.5$$ millimole and $$E = 1.067$$ V, then calculate the pH of the solution.

Assertion (A) Sodium acetate on Kolbe’s electrolysis gives ethane.

Reason (R) Methyl free radical is formed at cathode.