$$\eqalign{ & {E_{\frac{{Cu}}{{C{u^{2 + }}}}}} = E_{\frac{{Cu}}{{C{u^{2 + }}}}}^ \circ - \frac{{0.0591}}{2}\log \left[ {C{u^{2 + }}} \right] \cr & {\text{If log}}\left[ {C{u^{2 + }}} \right] = 0\,\,{\text{i}}{\text{.e}}.\left[ {C{u^{2 + }}} \right] = 1 \cr & {\text{then}}\,\,{E_{\frac{{Cu}}{{C{u^{2 + }}}}}} = E_{\frac{{Cu}}{{C{u^{2 + }}}}}^ \circ \cr & OA = 0.34\,\,V = E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ = - E_{\frac{{Cu}}{{C{u^{2 + }}}}}^ \circ \cr & \Rightarrow E_{\frac{{Cu}}{{C{u^{2 + }}}}}^ \circ = - 0.34\,V \cr & {\text{Now,}}\,{E_{\frac{{Cu}}{{C{u^{2 + }}}}}} = - 0.34 - \frac{{0.0591}}{2}\log \,0.1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 0.34 + \frac{{0.0591}}{2}V \cr} $$

$$\eqalign{ & \Delta G\,{\text{of}}\,{H_2}O\left( l \right) = - 237.2\,kJ/mol \cr & \Delta G\,{\text{of}}\,C{O_2}\left( g \right) = - 394.4\,kJ/mol \cr & \Delta G\,{\text{of pentane}}\left( g \right) = - 8.2\,kJ/mol \cr} $$
In pentane-oxygen fuel cell following reaction takes place
$$\eqalign{ & \,\,\,{C_5}{H_{12}} + 10{H_2}O\left( l \right) \to 5C{O_2} + 32{H^ + } + 32{e^ - } \cr & 8{O_2} + 32{H^ + } + 32{e^ - } \to 16{H_2}O\left( l \right) \cr & \overline {{C_5}{H_{12}} + 8{O_2} \to 5C{O_2} + 6{H_2}O\left( l \right),{E^ \circ } = ?} \cr & \Delta {G_{{\text{reaction}}}} = \sum {\Delta {G_{{\text{product}}}} - \sum {\Delta {G_{{\text{reactant}}}}} } \cr} $$
  $$ = 5 \times \Delta {G_{\left( {C{O_2}} \right)}} + 6\Delta {G_{\left( {{H_2}O} \right)}} - $$      $$\left[ {\Delta {G_{\left( {{C_5}{H_{12}}} \right)}} + 8 \times \Delta {G_{{O_2}}}} \right]$$
  $$ = 5 \times \left( { - 394.4} \right) + 6 \times \left( { - 237.2} \right)$$      $$ - \left( { - 8.2 + 0} \right)$$
$$\eqalign{ & = - 1972 - 1423.2 + 8.2 \cr & = - 3387\,kJ/mol \cr & = - 3387 \times {10^3}\,J/mol \cr & \Delta G = - nFE_{cell}^ \circ \cr & - 3387 \times {10^3} = - 32 \times 96500 \times E_{cell}^ \circ \cr & E_{cell}^ \circ = \frac{{ - 3387 \times {{10}^3}}}{{ - 32 \times 96500}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.0968\,V \cr} $$

As we go down the group 1( i.e. from $$L{i^ + }$$ to $${K^ + }$$ ), the ionic radius increases, degree of solvation decreases and hence effective size decreases resulting in increase in ionic mobility. Hence equivalent conductance at infinite dilution increases in the same order.

$${E_{{\text{oxidation}}}} = 0.059\,pH = 0.059 \times 10 = 0.59\,V$$

$$\eqalign{ & {\lambda _m} = \frac{{1000\kappa }}{{0.1}} \cr & = \frac{{1000 \times 3.75 \times {{10}^{ - 4}}}}{{0.1}} \cr & = 3.75; \cr & \alpha = \frac{{{\lambda _m}}}{{\lambda _m^\infty }} \cr & = \frac{{3.75}}{{250}} \cr & = 1.5 \times {10^{ - 2}}; \cr & {K_a} = C{\alpha ^2} \cr & = 0.1 \times {\left( {1.5 \times {{10}^{ - 2}}} \right)^2} \cr & = 2.25 \times {10^{ - 5}} \cr} $$

The given order of reduction potentials is $$Z > Y > X.$$    A spontaneous reaction will have the following characteristics
$$Z$$ reduced and $$Y$$ oxidised
$$Z$$ reduced and $$X$$ oxidised
$$Y$$ reduced and $$X$$ oxidised
Hence, $$Y$$ will oxidise $$X$$ and not $$Z.$$

Difluoroacetic acid being strongest acid will furnish maximum number of $$ions$$  showing highest electrical conductivity.

On pulling out one of the electrodes the electric field applied to the solution disappears and hence the ions start moving randomly.

Pure metal always deposits at cathode.

$$\eqalign{ & {\text{Correct Nernst equation is}} \cr & E = {E^ \circ } + \frac{{2.303\,RT}}{{nF}}\log {a_{{M^{n + }}}}. \cr} $$