Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

On charging the battery the reaction is reversed and $$PbS{O_{4\left( s \right)}}$$  is converted into $$Pb$$  at anode and $$Pb{O_2}$$  at cathode.

$$\eqalign{ & {\text{Volume = Area }} \times {\text{thickness}} \cr & {\text{Mass = Volume}}\,\, \times \,{\text{density}} \cr & \therefore \,\,{\text{Mass of }}Ag{\text{ to be deposited}} \cr & {\text{ = }}\frac{{80 \times 0.005}}{{10}} \times 10.5 \cr & = 0.42\,g \cr & {\text{Amount deposited}} = \frac{{i \times t \times E.wt}}{{96500}} \cr & \therefore \,\,0.42 = \frac{{108 \times 3 \times t}}{{98500}} \cr & \therefore \,\,t = 125.1\,{\text{seconds}} \cr} $$

Key Idea
$$\eqalign{ & {\lambda _{eq}} = \kappa \times V \cr & \,\,\,\,\,\,\,\, = \frac{{k \times 1000}}{{{\text{Normality}}}} \cr} $$
On dilution, the number of current carrying particles per $$c{m^3}$$ decreases but the volume of solution increases. Consequently, the ionic mobility increases, which in turn increases the equivalent conductance of strong electrolyte.

$$\underset{\left(+6\right)}{Mn{O}_4^{2-}}\stackrel{\left[1molee{}^{-}=1F\right]}{\rightleftharpoons }\underset{\left(+7\right)}{Mn{O}_4^{-}}$$
As per the equation, for $$1$$ $$mole$$  of $$MnO_4^{2 - },$$  $$1$$ $$F$$ of electricity is required. Thus, for $$0.1$$ $$mole$$  of $$MnO_4^{2 - },$$  $$0.1$$ $$F$$ of electricity is required.
Since, $$1F = 96500\,C$$
$$\therefore \,\,0.1\,F = 0.1 \times 96500\,C = 9650\,C$$
Hence, $$9650$$ $$C$$ of electricity is required to completely oxidise $$MnO_4^{2 - }$$  to $$MnO_4^ - .$$

Conductivity depends upon solvation of ions present in solution. Greater the solvation of ions, lesser is the conductivity.

Sodium stearate at low concentration ( i.e., below $$CMC$$ ) behaves as normal strong electrolyte, but at higher concentration ( i.e. above $$CMC$$ ) exhibits colloidal behaviour due to the formation of micelles. Thus, plot (D) correctly represents relation between $$^ \wedge m$$  and $$\sqrt C $$ for sodium strearate.

Degree of dissociation, $$\,\alpha = \frac{{{\Lambda ^c}}}{{{\Lambda ^\infty }}}$$
where $${{\Lambda ^c}}$$ and $${{\Lambda ^\infty }}$$ are equivalent conductances at a given concentration and at infinite dilution respectively.
$$\eqalign{ & \Rightarrow \alpha = \frac{{8.0}}{{400}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 2 \times {10^{ - 2}} \cr} $$
From Ostwald's dilution law ( for weak monobasic acid ),
$$\eqalign{ & {K_c} = \frac{{C{\alpha ^2}}}{{\left( {1 - \alpha } \right)}} \cr & {\text{or}}\,\, = C{\alpha ^2}\,\,\,\left( {\because \,\,1 > > > \alpha } \right) \cr & = \frac{1}{{32}}{\left( {2 \times {{10}^{ - 2}}} \right)^2} \cr & = 1.25 \times {10^{ - 5}} \cr} $$

Magnesium provides cathodic protection and prevent rusting or corrosion.

NOTE : According to Kohlrausch’s law, molar conductivity of weak electrolyte acetic acid $$\left( {C{H_3}COOH} \right)$$   can be calculated as follows :
$${\Lambda ^ \circ }_{C{H_3}COOH} = \left( {{\Lambda ^ \circ }_{C{H_3}COONa} + {\Lambda ^ \circ }_{HCl}} \right) - {\Lambda ^ \circ }_{NaCl}$$
∴ Value of $${\Lambda ^ \circ }_{NaCl}$$  Should also be known for calculating value of $${C{H_3}COOH}$$  .

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