Chemistry MCQs for NEET — Practice Questions with Answers

(d) 

$$\begin{gathered} E°=E + \frac{0.058}{1}\log {10}^{-7} \\ = E° + \frac{0.059 \times (-7)}{1} =E°- 0.41 \mathrm{V} \end{gathered}$$

(b) The charging of lead storage battery involves the reverse reactions.

The spontaneous direction of a galvanic cell reaction is from higher potential to lower potential. If P1 < P2, it means the potential of the left half-cell is lower than the right half-cell, so the reaction will proceed spontaneously from right to left.

Concept used: Faradays second law + electrolysis

$\frac{{\mathrm{w}}_1}{\mathrm{Eq} {\mathrm{wt}}_1}=\frac{w_2}{Eq w{t}_2}$

= Electrolysis of ${\mathrm{H}}_2{\mathrm{SO}}_4$

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-} \\ {\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{O}}^{2-} \end{gathered}$$

at cathode ${\mathrm{H}}_2$ ags will be deposit

electrolysis of ${\mathrm{mgSO}}_4$

$$\begin{gathered} {\mathrm{mgSO}}_4\to {\mathrm{mg}}^{2+}+{\mathrm{SO}}_4^{2-} \\ {\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{H}}^{+}+ {\mathrm{O}}^{2-} \end{gathered}$$

at cathode: $H_2$ gas will evolved because ${\mathrm{H}}^{+}$ ions has lower discharge potential. than ${\mathrm{mg}}^{2+}$

hence, mg will not get deposited

(a) The equivalent conductance of BaCl₂ at infinite dilution,

${\lambda }_{\infty }$ of BaCl₂ = (1/2)${\lambda }_{\infty }$of Ba²⁺ + ${\lambda }_{\infty }$ of Cl^-

= 127/2+76 = 139.5

(b) MnO₂ in Lechlanche cell.

Only oxidation occurs at anode. Also discharge potential of H₂O is less than discharge potential of ${\mathrm{SO}}_4^{2-}$

(a) ${\mathrm{\lambda }}_{\mathrm{eq}} = \mathrm{k} \times \mathrm{V} = \frac{\mathrm{k} \times 1000}{\mathrm{Normality}}$

On dilution, the number of current carrying particles per cm³ decreases but the volume of solution increases. Consequently, the ionic mobility increases, which in turn increases the equivalent conductance of strong electrolyte.

Concept used: nearest equation

$\mathrm{Tl}/ {\mathrm{Tl}}^{+} (0.001\mathrm{M})/\mathrm{l} {\mathrm{Cu}}^{2+} (0.1 \mathrm{M})/ \mathrm{Cu}$

     $$\begin{gathered} \mathrm{Ti} \to {\mathrm{Ti}}^{+}+{\mathrm{e}}^{-}] \times 2 \\ {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu} \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ 2\mathrm{Ti} + {\mathrm{Cu}}^{2+} \to 2{\mathrm{Ti}}^{+}+\mathrm{Cu} \end{gathered}$$

for this cell nearest equation:

${\mathrm{E}}_{\mathrm{cell}}-{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.059}{\mathrm{n}}10\mathrm{g} \frac{{\left[{\mathrm{Ti}}^{+}\right]}^2}{\left[{\mathrm{Cu}}^{2+}\right]}$

if ${\mathrm{Cu}}^{2+}$ ion cone $\uparrow \uparrow \uparrow$ then:

$\frac{0.059 }{n} \log \frac{{\left[T{i}^{+}\right]}^2}{C{u}^{2+}}\downarrow \downarrow \downarrow$

but overall ${\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.059}{\mathrm{n}}$ increases

hence ${\mathrm{E}}_{\mathrm{cell}}$ increases

So on increasing ${\mathrm{Cu}}^{2+}$ ion conc. 

${\mathrm{E}}_{\mathrm{cell}}$ increases.