Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$Zn\left( s \right) + Cu_{\left( {aq} \right)}^{2 + } \to Zn_{\left( {aq} \right)}^{2 + } + Cu\left( s \right)$$
$$\Delta G = \Delta {G^o} + 2.303\,RT\,\,{\log _{10}}Q;$$       $$Q = \frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {C{u^{2 + }}} \right]}}$$
$$\left[ {\Delta {G^o} = - nF{E^o}} \right] = - 2 \times F \times 1.1$$
$${\text{Given}}\left[ {Z{n^{2 + }}} \right] = 10\left[ {C{u^{2 + }}} \right]$$
$$\therefore \,\,\Delta G = - 2F\left( {1.1} \right) + 2.303RT\,{\text{lo}}{{\text{g}}_{{\text{10}}}}10$$       $$ = 2.303RT - 2.2F$$

$$\eqalign{ & {\text{Conductivity}}\left( \kappa \right) = \frac{\ell }{{R.a.}} \cr & \kappa = \frac{l}{{R.a}};\,{\text{cell constant}}.\left( {\frac{l}{a}} \right) = 1.29 \times 100 = 129 \cr & {\text{Again conductivity of }}0.02M{\text{ solution}} \cr & \kappa {\text{ = }}\frac{1}{{520}} \times 129 \cr & { \wedge _m} = \frac{{\kappa \times 1000}}{M} \cr & = \frac{{129}}{{520}} \times \frac{{1000}}{{0.02}} \cr & = 1.24 \times {10^{ - 4}}S\,{m^2}\,mo{l^{ - 1}} \cr} $$

$$2N{H_4}Cl + Zn \to 2N{H_3} + ZnC{l_2} + {H_2}\, \uparrow .$$

According to the definition $$1\,F$$  or $$96500\,C$$  is the charge carried by $$1\,mol$$  of electrons when water is electrolysed
$$2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }$$
So, 4 Faraday of electricity liberate $$ = 32\,g$$  of $${O_2}.$$
Thus 1 Faraday of electricity liberate
$$ = \frac{{32}}{4}g$$   of $${O_2} = 8\,g$$   of $${O_2}$$

Lower the value of reduction potential higher will be reducing power hence the correct order will be $$M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$$

$$\eqalign{ & \because \,\,Q = i \times t \cr & \therefore \,\,Q = 4.0 \times {10^4} \times 6 \times 60 \times 60C \cr & = 8.64 \times {10^8}C \cr & {\text{Now since }}96500{\text{ }}C{\text{ liberates }}9{\text{ }}g{\text{ of }}Al \cr & 8.64 \times {10^8}\,C\,{\text{liberates}} \cr & \frac{9}{{96500}} \times 8.64 \times {10^8}g\,Al = 8.1 \times {10^4}g\,{\text{of}}\,Al \cr} $$

$$\eqalign{ & A{l_2}{O_3} \to 2A{l^{3 + }} + 3{O^{2 - }} \cr & 2A{l^{3 + }} + 6{e^ - } \to 2Al \cr & \therefore A{l^{3 + }} + 3{e^ - } \to Al \cr} $$
Thus, 3 $$F$$  of charge is required to obtain $$1\,mole$$  of $$Al$$  from $$A{l_2}{O_3}.$$

These are not electrolytes.

$$\eqalign{ & {\text{For spontaneous}} \cr & {\Delta _r}{G^ \circ } = - nF{E^ \circ } < 0 \cr & \therefore \,\,{E^ \circ } > 0 \cr} $$

$$E_{RP\left( {\frac{{C{r_2}O_7^{2 - }}}{{C{r^{3 + }}}}} \right)}^ \circ = 1.33\,V$$     and $$E_{RP\left( {\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}} \right)}^ \circ = 0.77\,V$$
$$\therefore \,E_{RP\left( {\frac{{C{r_2}O_7^{2 - }}}{{C{r^{3 + }}}}} \right)}^ \circ $$     is more positive than $$E_{RP\left( {\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}} \right)}^ \circ $$
thus, $$C{r_2}O_7^{2 - }$$  will undergo reduction and $$F{e^{2 + }}$$  will undergo oxidation so. electrons will flow from $$Fe$$  electrode to $$Cr$$  electrode. Also. $$Fe$$  electrode will be negative electrode. Also.
$$\eqalign{ & E_{{\text{cell}}}^ \circ = E_{RP\left( {\frac{{C{r_2}O_7^{2 - }}}{{C{r^{3 + }}}}} \right)}^ \circ - E_{RP\left( {\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}} \right)}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.33 - 0.77 \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.56\,V \cr} $$