Electrochemistry

$1$ mole of ${e^ - } = 1F = 96500\,C$

$27g$ of $Al$ is deposited by $3 \times 96500\,C$

$5120$ $g$ of $Al$ will be deposited by

$ = {{3 \times 96500 \times 5120} \over {27}}$

$ = 5.49 \times {10^7}C$

$E_{cell}^o = E_{cell}^o - {{0.059} \over n}\log \,{K_c}$

or $\,\,\,\,\,\,$ $0 = 0.591 - {{0.0591} \over 1}\log {K_c}$

or $\,\,\,\,\,\,$ $\log \,{K_c} = {{0.591} \over {0.0591}} = 10$

or $\,\,\,\,\,\,$ ${K_c} = 1 \times {10^{10}}$

$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\Delta {G^ \circ }$

$\,\,\,\,\,\,\,\,\,\, = - 1 \times F \times 0.77$

$S{n^{2 + }} + 2{e^ - } \to Sn\left( s \right)\Delta {G^ \circ }$

$ = - 2 \times F\left( { - 0.14} \right)$

$Sn\left( s \right) + 2F{e^{3 + }}\left( {aq} \right)$

$\,\,\,\,\,\,\,\,\,\, \to 2F{e^{2 + }}\left( {aq} \right) + S{n^{2 + }}\left( {aq} \right)$

$\therefore$ Standard potential for the given reaction

or $E_{cell}^o = E_{Sn/S{n^{2 + }}}^o + E_{F{e^{3 + }}/F{e^{2 + }}}^0$

$ = 0.14 + 0.77 = 0.91\,V$

$Zn\left( s \right) + 2{H^ + } + \left( {aq} \right)\,\,\rightleftharpoons\,\,Z{n^{2 + }}\left( {aq} \right) + {H_2}\left( g \right)$

${E_{cell}} = E_{cell}^ \circ - {{0.059} \over 2}\log {{\left[ {Z{n^{2 + }}} \right]\left[ {{H_2}} \right]} \over {{{\left[ {{H^ + }} \right]}^2}}}$

Addition of ${H_2}S{O_4}$ will increases $\left[ {{H^ + }} \right]$ and ${E_{cell}}$ will also increase and the equilibrium will shift towards $RHS$

The given values show that $Cr$ has maximum oxidation potential, therefore its oxidation will be easiest. (Change the sign to get the oxidation values)

${A^ \circ }Nacl = {\lambda ^ \circ }N{a^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

${A^ \circ }KBr = {\lambda ^ \circ }{K^ + } + \lambda Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

${A^ \circ }KCl = {\lambda ^ \circ }{K^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$

operating $(i)+(ii)-(iii)$

${A^ \circ }NaBr = {\lambda ^ \circ }N{a^ + } + {\lambda ^ \circ }B{r^ - }$

$ = 126 + 152 - 150$

$ = 128\,\,\,S\,c{m^2}\,mo{l^{ - 1}}$

In ${H_2} - {O_2}$ fuel cell, the combustion of ${H_2}$ occurs to create potential difference between the two electrodes

The equlibrium constant is released to the standard emf of cell by the expression

$\log \,K = {E^ \circ }_{cell} \times {n \over {0.059}}$

$ = 0.295 \times {2 \over {0.059}}$

$\log \,K = {{590} \over {59}} = 10$

or $\,\,\,\,\,K = 1 \times {10^{10}}$

${E_{cell}} = {E^ \circ }_{cell} + {{0.059} \over n}\log {{\left[ {C{u^{ + 2}}} \right]} \over {\left[ {Z{n^{ + 2}}} \right]}}$

$ = 1.10 + {{0.059} \over 2}\log \left[ {0.1} \right]$

$ = 1.10 - 0.0295$

$ = 1.07V$

Magnesium provides cathodic protection and prevent rusting or corrosion.

$\matrix{ A & B & C \cr { + 0.5C} & { - 3.0V} & { - 1.2V} \cr } $

NOTE : The higher the negative value of reduction potential, the more is the reducing power.

Hence $B > C > A.$

When $96500$ coulomb of electricity is passed through the electroplating bath the amount of Ag deposited $=108g$

$\therefore$ when $9650$ coulomb of electricity is passed deposited Ag.

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{108} \over {96500}} \times 9650 = 10.8\,g$

Given $S \propto {{area\, \times \,conc} \over \ell } = {{\kappa {m^2}mol} \over {m \times {m^3}}}$

$\therefore$ $\,\,\,\,\kappa = S{m^2}mo{l^{ - 1}}$

Oxidation half cell : -

${H_2}\left( g \right)\buildrel \, \over \longrightarrow 2{H^ + }\left( {1M} \right) + 2{e^ - }\,\,\,...{P_1}$

Reduction half cell

$2{H^ + }\left( {1M} \right) + 2{e^ - }\buildrel \, \over \longrightarrow {H_2}\left( g \right)\,\,\,...{P_2}$

The net cell reaction

$\mathop {{H_2}}\limits_{{P_1}} \left( g \right)\buildrel \, \over \longrightarrow \mathop {{H_2}}\limits_{{P_2}} \left( g \right)$

$E_{cell}^ \circ = 0.00\,V$ $\,\,\,\,\,$ $n=2$

$\therefore$ $\,\,\,\,$ ${E_{cell}} = E_{cell}^ \circ - {{RT} \over {nF}}{\log _c}K$

$ = 0 - {{RT} \over {nF}}{\log _e}{{{P_2}} \over {{P_1}}}$

or $\,\,\,\,{E_{cell}} = {{RT} \over {2F}}{\log _e}{{{P_1}} \over {{P_2}}}$

Pure metal always deposits at cathode.

$2C{r^{3 + }} + 7{H_2}O \to C{r_2}O_7^{2 - } + 14{H^ + }$

$O.S.\,\,$ of $Cr$ changes from $+3$ to $+6$ by loss of electrons.

At anode oxidation takes place.

${E_{cell}} = \,\,$ Reduction potential of cathode (right)

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $-$ Reduction potential of anode (left)

$ = {E_{right}} - {E_{left}}.$

The cell notation indicates that the left side (Ag|Ag+) is the anode (oxidation occurs), and the right side (Cu²⁺|Cu) is the cathode (reduction occurs). In a galvanic (voltaic) cell, electrons flow from anode to cathode.

The standard cell potential (E° cell) is the difference in potential between the cathode and anode. It's calculated as :

E°_cell = E°_cathode - E°_anode

At LHS (oxidation) :

$ 2 \times\left(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-}\right); \quad E_{\text {oxi }}^{\circ}=-x $

At RHS (reduction) :

$ \mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}; \quad \mathrm{E}_{\text {red }}^{\circ}=+y $

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$ 2 \mathrm{Ag}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}+2 \mathrm{Ag}^{+}, E_{\text {cell }}^{\circ}=(y-x) $
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$ \therefore $ E°_cell = y - x

Therefore, the correct answer is :

Option C : y - x.