Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

In the aqueous solution, if $${H_2}S{O_4}$$  is present then iron will form hydrated iron sulphate rather than forming $${\left[ {FeS{O_4}} \right]^ + }$$  solution. Moreover, tendency of formation of $${FeS{O_4}}$$  or $$F{e_2}{\left( {S{O_4}} \right)_3}$$   is even more. Formation of $${\left[ {FeHP{O_4}} \right]^ + }$$   is reasonably fine. Due to the formation of complex. $$\left[ {F{e^{3 + }}} \right]$$  decreases and accordingly. potential decreases.
$$E = {E^ \circ } - \frac{{0.0591}}{n}{\log _{10}}\frac{{\left[ {F{e^{2 + }}} \right]}}{{\left[ {F{e^{3 + }}} \right]}}$$

Oxidation takes place at magnesium electrode and reduction at hydrogen electrode.

The value of $$E_{_{\frac{{{M^{2 + }}}}{M}}}^ \circ $$  for given metal ions are
$$\eqalign{ & E_{_{\frac{{M{n^{2 + }}}}{{Mn}}}}^ \circ = - 1.18\,V,\,\,E_{\frac{{C{r^{2 + }}}}{{Cr}}}^ \circ = - 0.9\,V, \cr & E_{_{\frac{{F{e^{2 + }}}}{{Fe}}}}^ \circ = - 0.44\,V\,\,{\text{and}}\,\,E_{_{\frac{{C{o^{2 + }}}}{{Co}}}}^ \circ = - 0.28\,V. \cr} $$
The correct order of $$E_{_{\frac{{{M^{2 + }}}}{M}}}^ \circ $$  values without considering negative sign would be $$M{n^{2 + }} > C{r^{2 + }} > F{e^{2 + }} > C{o^{2 + }}.$$

$$\eqalign{ & C{u^{2 + }} + 2{e^ - } \to Cu \cr & 2F\,i.e.\,\,2 \times 96500\,C\,deposit\,Cu = 1\,mol = 63.5\,g \cr} $$

$${\text{Given that}}\,{\text{E}}_{\frac{{F{e^{2 - }}}}{{Fe}}}^ \circ = - 0.441\,V$$
$${\text{So,}}\,\,Fe \to F{e^{2 + }} + 2{e^ - },$$     $${E^ \circ } = + 0.441\,V\,...{\text{(i)}}$$
$${\text{and}}\,\,\,\,E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^ \circ = 0.771\,V$$
$${\text{So,}}\,2\,F{e^{3 + }} + 2{e^ - } \to 2\,F{e^{2 + }},$$      $${E^ \circ } = 0.771\,V\,\,...{\text{(ii)}}$$
$$\eqalign{ & {\text{Cell reaction}} \cr & \left( {\text{i}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Fe \to F{e^{2 + }} + 2{e^ - },{E^ \circ } = 0.441\,V \cr & \underline {\left( {{\text{ii}}} \right)2F{e^{3 + }} + 2{e^ - } \to 2F{e^{2 + }},\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = + 0.771\,V} \cr & \underline {\,\,\,\,\,\,\,Fe + 2F{e^{3 + }} \to 3F{e^{2 + }},\,\,\,\,\,\,\,\,\,\,\,E_{{\text{cell}}}^ \circ = 1.212\,V} \cr} $$
Alternative On the basis of cell reaction following half-cell reactions are written
$$\eqalign{ & {\text{At anode}} \cr & Fe \to F{e^{2 + }} + 2{e^ - }\,{\text{(oxidation)}} \cr & {\text{At cathode}} \cr & {\text{2F}}{{\text{e}}^{3 + }} + 2{e^ - } \to 2F{e^{2 + }}\,\,{\text{(reduction)}} \cr & {\text{So,}}\,{\text{E}}_{{\text{cell}}}^ \circ = {\text{E}}_{{\text{cathode}}}^ \circ - {\text{E}}_{{\text{anode}}}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( { + 0.771} \right) - \left( { - 0.441} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 1.212\,V. \cr} $$

$$\eqalign{ & {\text{Oxidising tendency}} \propto \frac{1}{{{\text{Electrode potential}}}} \cr & TX \to {\text{No reaction}} \cr & TY \to X,Z \cr & TZ \to X \cr & \Rightarrow {\text{order of electrode potential is}} \cr & TY < TZ < TX \cr & \Rightarrow {\text{Order of oxidation of the anion is}} \cr & {Y^ - } > {Z^ - } > {X^ - } \cr} $$

Conductivity $$(X) =$$  conductance $$\left( c \right) \times $$  cell constant
∴ Cell constant $$ = \frac{X}{Y}$$
Conductivity of $$NaOH = \frac{X}{Y}.Z$$
Molar conductance of $$NaOH$$
$$\eqalign{ & = \frac{X}{Y}.Z \times \frac{{1000}}{{0.1}} \cr & = \frac{{XZ}}{Y}{10^4} \cr} $$

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Oxidation potential of Co is more than $$Ag,$$  hence cell reaction will be
$$Co + 2A{g^ + } \to C{o^{2 + }} + 2Ag$$
$${E_{cell}} = E_{cell}^ \circ - \frac{{RT\,ln}}{{nF}}\frac{{\left[ {C{o^{2 + }}} \right]}}{{{{\left[ {A{g^ + }} \right]}^2}}},$$       the lesser the value of the factor $$\frac{{\left[ {C{o^{2 + }}} \right]}}{{{{\left[ {A{g^ + }} \right]}^2}}}$$   greater will be value of $${E_{cell\, \cdot }}$$