$$P{t^{4 + }} + 4{e^ - } \to Pt$$
$$4\,moles$$  or $$4\,F$$  of electricity is required to deposit $$1\,mole$$  of $$Pt.$$
$$0.80\,F$$  of electricity will deposit $$\frac{1}{4} \times 0.80 = 0.20\,mol$$

$${\text{At cathode}}:Ag_{\left( {aq} \right)}^ + + {e^ - } \to A{g_{\left( s \right)}}$$
$${\text{At anode :}}\,{\text{2OH}}_{\left( {aq} \right)}^ - \to \frac{1}{2}{O_{2\left( g \right)}}$$       $$ + {H_2}{O_{\left( l \right)}} + 2{e^ - }$$

Due to buffer action the pH will remain practically constant.

$$\eqalign{ & {\text{Equivalent conductivity of}}\,BaC{l_2} \cr & \wedge _m^\infty \left( {BaC{l_2}} \right) = \lambda _m^\infty \left( {B{a^{2 + }}} \right) + 2\lambda _m^\infty \left( {C{l^ - }} \right) \cr & = 127 + 2 \times 76 = 279\,oh{m^{ - 1}}\,c{m^2}\,e{q^{ - 1}}. \cr} $$

As reduction potential $$Y$$ is higher than $$X$$ but lower than $$Z$$ thus $$Y$$ will oxidise $$X$$ and not $$Z.$$

From the question, we have an equation
$$\eqalign{ & 2{H^ + } + 2{e^ - } \to {H_2}\left( g \right) \cr & {\text{According to Nernst equation,}} \cr & E = {E^ \circ } - \frac{{0.0591}}{2}{\text{log}}\frac{{{p_{{H_2}}}}}{{{{\left[ {{H^ + }} \right]}^2}}} \cr} $$
$$ = 0 - \frac{{0.0591}}{2}{\text{log}}\frac{{{p_{{H_2}}}}}{{{{\left( {{{10}^{ - 7}}} \right)}^2}}}$$       $$\left[ {\because \,\,\left[ {{H^ + }} \right] = {{10}^{ - 7}}} \right]$$
∴ For potential of $${H_2}$$  electrode to be zero $${{p_{{H_2}}}}$$  should be equal to $${\left[ {{H^ + }} \right]^2},$$  i.e. $${10^{ - 14}}atm.$$
$$\therefore \,{\text{log}}\frac{{{{10}^{ - 14}}}}{{{{\left( {{{10}^{ - 7}}} \right)}^2}}} = 0$$

No explanation is given for this question. Let's discuss the answer together.

$$Zn\left( s \right) + C{u^{2 + }}\left( {aq} \right) \rightleftharpoons $$     $$Cu\left( s \right) + Z{n^{2 + }}\left( {aq} \right),{E^ \circ } = + 1.10\,V$$
$$\eqalign{ & \therefore \,\,{E^ \circ } = \frac{{0.0591}}{n}{\text{lo}}{{\text{g}}_{10}}\,{K_{eq}} \cr & {\text{because at equilibrium}},\,{E_{cell}} = 0 \cr & \left( {n = \,{\text{number of electrons exchanged}} = 2} \right) \cr & \,\,\,\,\,\,\,1.10 = \frac{{0.0591}}{2}{\text{lo}}{{\text{g}}_{10}}\,{K_{eq}} \cr & \frac{{2.20}}{{0.0591}} = {\text{lo}}{{\text{g}}_{10}}\,{K_{eq}} \cr & \,\,\,\,\,\,\,\,{K_{eq}} = {\text{antilog}}\,\,37.225 \cr & \,\,\,\,\,\,\,\,{K_{eq}} = 1.66 \times {10^{ - 37}} \cr} $$

No explanation is given for this question. Let's discuss the answer together.

Higher the value of reduction potential stronger will be the oxidising hence based on the given values $$A{g^ + }$$  will be strongest oxidizing agent.