$$\eqalign{ & {\text{From Faraday's}}\,{\text{first law of electrolysis,}} \cr & \frac{w}{E} = \frac{{it}}{{96500}}\,\,\,...{\text{(i)}} \cr & {\text{Given,}}\,i = 1\,A;\,t = 60\,s \cr & {\text{Putting these values in Eq}}.{\text{(i), we get}} \cr & \frac{w}{E} = \frac{{1 \times 60}}{{96500}}\,\,{\text{or}}\,\,\frac{w}{E} = \frac{6}{{9650}} \cr & = {\text{Number of mole of electrons}} \cr & \therefore \,\,{\text{Number of electrons}} \cr & = \frac{6}{{9650}} \times 6.022 \times {10^{23}} \cr & = 3.75 \times {10^{20}} \cr} $$

Ionic molar conductivity of $${H^ + }$$  is very high and $$N{H_4}OH$$   is a weak electrolyte.

Higher the reduction potential, stronger is the oxidising agent.

$${\text{Give}}:\,I = 10\,{\text{milliamperes ;}}$$      $$IF = 96500{\text{ }}C{\text{ }}mo{l^{ - 1}}$$
$$\eqalign{ & t = ?\,;\,{\text{Moles of}}\,{H_2}{\text{produces = 0}}{\text{.01 }}mol \cr & {\text{From the law of electrolysis, we have }} \cr & {\text{Equivalents of}}\,{H_2}\,{\text{produces}} = \frac{{I \times t\left( {\sec } \right)}}{{96500}} \cr & {\text{Substituting given values, we get}} \cr & 0.01 \times 2 = \frac{{10 \times {{10}^{ - 3}}\left( {{\text{amperes}}} \right) \times t\left( {\sec } \right)}}{{96500}} \cr & {\text{or}}\,\,t = \frac{{0.01 \times 2 \times 96500}}{{10 \times {{10}^{ - 3}}}}\sec \cr & \,\,\,\,\,\,\,\,\,\,\, = 19.3 \times {10^4}\sec . \cr & {\text{i}}{\text{.e}}{\text{. (B) is the correct answer}}{\text{. }} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & F{e^{3 + }} + 3{e^ - } \to Fe\,;\,E_1^ \circ = - 0.036\,V. \cr & F{e^{2 + }} + 2{e^ - } \to Fe\,;\,E_2^ \circ = - 0.439\,V. \cr & {\text{Required}}\,{\text{equation}}\,{\text{is}} \cr & {\text{F}}{{\text{e}}^{3 + }} + {e^ - } \to F{e^{2 + }}\,\,;\,\,E_3^ \circ = ? \cr & {\text{Applying}}\,\Delta {G^ \circ } = - nF{E^ \circ } \cr & \therefore \,\Delta G_3^ \circ = \Delta G_1^ \circ - \Delta G_2^ \circ \cr & \left( { - {n_3}FE_3^ \circ } \right) = \left( { - {n_1}FE_1^ \circ } \right) - \left( { - {n_2}FE_2^ \circ } \right) \cr & E_3^ \circ = 3E_1^ \circ - 2E_2^ \circ = 3 \times \left( { - 0.036} \right) - 2 \times \left( { - 0.439} \right) \cr & E_3^ \circ = - 0.108 + 0.878 = 0.77\,V. \cr} $$

Specific conductance $$ = S\,{m^{ - 1}}$$

$$\eqalign{ & \wedge _{m\,\,CaC{l_2}}^ \circ = \lambda _{C{a^{2 + }}}^ \circ + 2\lambda _{C{l^ - }}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 119.0 + 2 \times 76.3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 271.6\,S\,c{m^2}\,mo{l^{ - 1}} \cr & \wedge _{m\,\,C{H_3}COONa}^ \circ = \lambda _{C{H_3}CO{O^ - }}^ \circ + \lambda _{N{a^ + }}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 40.9 + 50.1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 91\,S\,c{m^2}\,mo{l^{ - 1}} \cr & \wedge _{m\,\,NaCl}^ \circ = \lambda _{Na}^ \circ + \lambda _{C{l^ - }}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 50.1 + 76.3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 126.4\,S\,c{m^2}\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & NaCl \to N{a^ + } + C{l^ - } \cr & 2C{l^ - } + 2{e^ - } \to C{l_2} \cr} $$
$$Q = I \times t = 2 \times 30 \times 60 = 3600$$       Coulombs
$$2 \times 96500$$   Coulombs produce $$ = 71\,g$$  of $$C{l_2}$$
3600 Coulombs will produce $$ = \frac{{71}}{{2 \times 96500}} \times 3600 = 1.32\,g$$       of $$C{l_2}$$

Higher the reduction potential, stronger is the oxidising agent.

Galvanization is the process by which zinc is coated over corrosive (easily rusted) metals to prevent them from corrosion.