Redox Reaction MCQ Questions & Answers in Physical Chemistry | Chemistry

$$N{a_2}\left[ {\underline {Fe} {{\left( {CN} \right)}_5}NO} \right]:$$     $$ + 2 + x + 5\left( { - 1} \right) + 0 = 0$$      $$ \Rightarrow x = + 3$$
$${K_2}\underline {Ta} {F_7}: + 2 + x + \left( { - 7} \right) = 0$$      $$ \Rightarrow x = + 5$$
$$M{g_2}{\underline P _2}{O_7}: + 4 + 2x + \left( { - 14} \right)$$       $$ = 0 \Rightarrow x = + 5$$
$$N{a_2}{\underline S _4}{O_6}: + 2 + 4x + \left( { - 12} \right)$$      $$ = 0\,\,{\text{or}}\,\,4x = 10 \Rightarrow x = + 2.5$$
$${\underline N _3}H:3x + 1 = 0 \Rightarrow x = - \frac{1}{3}$$

$$C{H_2}C{l_2} \to x + 2\left( { + 1} \right) + 2\left( { - 1} \right)$$       $$ = 0 \Rightarrow x = 0$$

Zinc gives $${H_2}$$  gas with dil $${H_2}S{O_4}/HCl$$    but not with $$HN{O_3}$$  because in $$HN{O_3},NO_3^ - \,ion$$    is reduced and give $$N{H_4}N{O_3},{N_2}O,NO$$     and $$N{O_2}$$  ( based upon the concentration of $$HN{O_3}$$  )
$$\eqalign{ & \left[ {Zn + \mathop {2HN{O_3}}\limits_{{\text{(nearly 6% )}}} \to Zn{{\left( {N{O_3}} \right)}_2} + 2H} \right] \times 4 \cr & HN{O_3} + 8H \to N{H_3} + 3{H_2}O \cr & N{H_3} + HN{O_3} \to N{H_4}N{O_3} \cr} $$
$$4Zn + 10HN{O_3} \to $$     $$4Zn{\left( {N{O_3}} \right)_2} + N{H_4}N{O_3} + 3{H_2}O$$
$$Zn$$  is on the top position of hydrogen in electrochemical series. So $$Zn$$  displaces $${H_2}$$  from dilute $${H_2}S{O_4}$$  and $$HCl$$  with liberation of $${H_2}.$$
$$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}$$

Since oxidation potential of $$Zn$$  is highest hence strongest reducing agent.

$$\eqalign{ & SO_3^{2 - } \to x + \left( { - 6} \right) = - 2 \Rightarrow x = + 4 \cr & SO_4^{2 - } \to x + \left( { - 8} \right) = - 2 \Rightarrow x = + 6 \cr & {S_2}O_4^{2 - } \to 2x + \left( { - 8} \right) = - 2 \Rightarrow x = + 3 \cr & {S_2}O_6^{2 - } \to 2x + \left( { - 12} \right) = - 2 \Rightarrow x = + 5 \cr & {\text{Hence}}\,\,SO_4^{2 - } > {S_2}O_6^{2 - } > SO_3^{2 - } > {S_2}O_4^{2 - } \cr} $$

The redox couple with maximum reduction potential will be best oxidising agent and with minimum reduction potential will be best reducing agent.

$$\eqalign{ & {\text{Given}}\,\,F{e^{3 + }}/F{e^{2 + }} = + 0.77\,V \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_2}/2{I^ - } = 0.536\,V \cr & 2\left( {{e^ - } + F{e^{3 + }} \to F{e^{2 + }}} \right)\,\,\,\,{E^ \circ } = 0.77\,V \cr & \underline {2{I^ - } \to {I_2} + 2{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = - 0.536\,V} \cr & 2F{e^{3 + }} + 2{I^ - } \to 2F{e^{2 + }} + {I_2} \cr & {E^ \circ } = {E^ \circ }_{ox} + {E^ \circ }_{red} \cr & = 0.77 - 0.536 \cr & = 0.164\,V \cr} $$
∴ Since value of $${E^ \circ }$$  is $$ + ve$$  reaction will take place.

$$2{I^ - } \to {I_2}$$   is oxidation (loss of electrons) ; $$Cr\left( { + 6} \right)$$   changes to $$Cr\left( { + 3} \right)$$  by gain of electrons. Hence $$Cr$$  is reduced.

Chromium does not show negative oxidation state.

$$\mathop {}\limits_{{E^ \circ }\left( V \right)} \,\,\,\mathop {Na}\limits_{ - 2.71} > \mathop {Mg}\limits_{ - 2.36} > \mathop {Zn}\limits_{ - 0.76} > \mathop {Fe}\limits_{ - 0.44} > \mathop {Cu}\limits_{ + 0.34} > \mathop {Ag}\limits_{ + 0.80} $$