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 $4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_2$
In an aqueous solution how does specific conductivity of electrolytes change with addition of water?
Conductivity of solution due to total ions present in per unit volume of solution is known as specific conductivity. Specific conductivity decreases due to decrease in the number of ions per unit volume. We add water to aqueous solution, number of ions present in per unit volume decreases.
Which reference electrode is used to measure the electrode potential of other electrodes?
Standard hydrogen electrode (SHE) is the reference electrode whose electrode potential is taken to be zero. The electrode potential of other electrodes are measured with respect to it .
Consider a cell given below
$$\mathrm{Cu}\left|\mathrm{Cu}^{2+}\right|\left|\mathrm{Cl}^{-}\right| \mathrm{Cl}_{2,}, \mathrm{Pt}$$
Write the reactions that occur at anode and cathode.
Cell reaction represented in the question is composed of two half cell reactions. These reactions are as follows
At anode $\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 e^{-}$
At cathode $\mathrm{Cl}_2+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}$
Copper is getting oxidised at anode. $\mathrm{Cl}_2$ is getting reduced at cathode.