Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

More negative is the value of reduction potential, higher will be the reducing property, i.e., the power to give up electrons.

In acidic medium, dichromate ions are reduced according to the equation, $$C{r_2}O_7^{2 - } + 14{H^ + } + 6{e^ - } \to $$       $$2C{r^{3 + }} + 7{H_2}O$$
$$1\,mole$$  of $$C{r_2}O_7^{2 - }$$  ion requires $$6\,moles$$  of electrons or $$6 \times 96500\,C$$   of electricity.

$$\eqalign{ & { \wedge _m} = \frac{{\kappa \times 1000}}{M} \cr & \,\,\,\,\,\,\,\,\, = \frac{{1.01 \times {{10}^{ - 2}} \times 1000}}{{0.1}} \cr & \,\,\,\,\,\,\,\,\, = 1.01 \times {10^2}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}} \cr} $$

Anode is always the site of oxidation thus, anode half-cell is
$$Z{n^{2 + }}\left( {aq} \right) + 2{e^ - } \to Zn\left( s \right),$$       $${E^ \circ } = - 0.76\,V$$
$${\text{Cathode half - cell is}}$$
$$A{g_2}O\left( s \right) + {H_2}O\left( l \right) + 2{e^ - } \to $$      $$2Ag\left( s \right) + 2O{H^ - }\left( {aq} \right),{E^ \circ } = 0.34\,V$$
$$\eqalign{ & {E^ \circ }_{{\text{cell}}} = {E^ \circ }_{{\text{cathode}}} - {E^ \circ }_{{\text{anode}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.34 - \left( { - 0.76} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 1.10\,V \cr} $$

$$\eqalign{ & {\text{(a)}}\,\,\,M{n^{2 + }} + 2{e^ - } \to Mn\,;\,{E^o} = - 1.18\,V;\,\,...{\text{(i)}} \cr & {\text{(b)}}\,\,\,M{n^{3 + }} + e \to M{n^{2 + }};\,{E^o} = - 1.51\,V;\,\,...{\text{(ii)}} \cr} $$
Now multiplying equation (ii) by two and subtracting from equation (i)
$${\text{3M}}{{\text{n}}^{2 + }} \to M{n^ + } + 2M{n^{3 + }};$$      $${E^o} = {E_{{\text{OX}}{\text{.}}}} + {E_{{\text{Red}}{\text{.}}}} = - 1.18 + \left( { - 1.51} \right)$$        $$ = - 2.69\,V$$
[ $$-ve$$  value of $$EMF$$  $$\left( {i.e.,\,\Delta G = + ve} \right)$$    shows that the reaction is non-spontaneous ]

In the silver plating of copper, $$K\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$   is used instead of $$AgN{O_3}.$$  Copper being more electropositive readily precipitate silver from their salt solution
$$Cu + 2AgN{O_3} \to Cu{\left( {N{O_3}} \right)_2} + Ag$$
whereas in $$K\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$   solution a complex anion $${\left[ {Ag{{\left( {CN} \right)}_2}} \right]^ - }$$  is formed and hence $$A{g^ + }$$ are less available in the solution and therefore copper cannot displace $$Ag$$  from its complex $$ion.$$

Molar conductivity is inversely proportional to molarity.

$$\eqalign{ & {\text{Given}}\,\,{E_{\frac{{S{n^{4 + }}}}{{S{n^{2 + }}}}}} = + 0.15\,V \cr & {E_{\frac{{C{r^{3 + }}}}{{Cr}}}} = - 0.74\,V \cr & E_{cell}^ \circ = E_{cathode}^ \circ - E_{anode}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.15 - \left( { - 0.74} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = + 0.89\,V \cr} $$

$$Ag$$  becomes oxidized and $${B{r^ - }}$$  becomes reduced.

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