A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell ?
When an opposing potential of 1.1 V is applied to a galvanic cell having electrical potential of 1.1 V then cell reaction stops completely and no current will flow through the cell.
How will the pH of brine (aq NaCl solution) be affected when it is electrolysed?
The pH of the solution will rise as NaOH is formed in the electrolytic cell. Chemical reaction occurring at cell when aqueous brine solution is electrolysed are as follows
$$\begin{aligned} & \mathrm{NaCl}(a q) \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q) \\ & \text { Cathode } \mathrm{H}_2 \mathrm{O}(l)+e^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2(g)+\mathrm{OH}^{-}(a q) \\ & \text { Anode Cl }(a q) \longrightarrow \frac{1}{2} \mathrm{Cl}_2(g)+e^{-} \\ \text{Net reaction}\quad& \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)+\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \end{aligned}$$
Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why ?
Life time of any cell depends upon ions present in cell. Ions are not involved in the overall cell reaction of mercury cell. Hence, mercury cell has a constant cell potential throughout its useful life.
Solutions of two electrolytes $A$ and $B$ are diluted. The $\Lambda_m$ of ' $B$ ' increases 1.5 times while that of $A$ increases 25 times. Which of the two is a strong electrolyte ? Justify your answer.
Strong electrolytes dissociate almost completely even on high concentration. Therefore, concentration of such solutions remain almost same on dilution. Electrolyte ' $B$ ' is stronger than ' $A$ ' because in ' $B$ ' the number of ions remains the same on dilution, but only interionic attraction decreases.
Therefore, $\Lambda_m$ increases only 1.5 times. While in case of weak electrolyte on dilution, number of constituent ions increases.
When acidulated water (dil. $\mathrm{H}_2 \mathrm{SO}_4$ solution) is electrolysed, with pH of the solution be affected ? Justify your answer.
Since pH of solution depends upon concentration of $\mathrm{H}^{+}$presence in solutions. pH of the solution will not be affected as $\left[\mathrm{H}^{+}\right]$remains constant.
At anode $2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-}$
At cathode $4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_2$