Redox Reaction MCQ Questions & Answers in Physical Chemistry | Chemistry

Oxidation number of hydrogen may be + 1 or - 1 or 0. For example, in $$HX$$  it is + 1, in hydrides $$(MH),$$  it is - 1 and in $${H_2}$$  it is 0.

Higher the reduction potential, stronger is oxidising agent.

At anode, oxidation takes place.

$$O.N.$$  of $$N$$  in $$N{O_2}$$  and $${N_2}{O_4}$$  is +4
∴ difference is zero.
$$O.N.$$  of $$P$$  in $${P_2}{O_5}$$  and $${{\text{P}}_4}{O_{10}}$$  is +5
∴ difference is zero
$$O.N.$$  of $$N$$  in $${N_2}O$$  is +1 and in $$NO$$  is +2. The difference is 1
$$O.N.$$  of $$S$$  in $$S{O_2}$$  is +4 and in $$S{O_3}$$  is +6. The difference is +2.

$${\text{The reaction given is}}$$
$$C{r_2}{O_7}^{2 - } + F{e^{2 + }} + {C_2}{O_4}^{2 - } \to $$       $$C{r^{3 + }} + F{e^{3 + }} + C{O_2}$$
$$C{r_2}{O_7}^{2 - } \to 2C{r^{3 + }}$$
$${\text{On balancing}}$$
$$14{H^ + } + C{r_2}{O_7}^{2 - } + 6{e^ - } \to $$      $$2C{r^{3 + }} + 7{H_2}O\,\,...\left( {\text{i}} \right)$$
$$\eqalign{ & F{e^{2 + }} \to F{e^{3 + }} + {e^ - }\,\,...\left( {{\text{ii}}} \right) \cr & {C_2}{O_4}^{2 - } \to 2C{O_2} + 2{e^ - }\,\,....\left( {{\text{iii}}} \right) \cr & {\text{On adding all the three equations}}. \cr} $$
$$C{r_2}{O_7}^{2 - } + F{e^{2 + }} + {C_2}O_4^{2 - } + 14{H^ + } + 3{e^ - }$$       $$ \to 2C{r^{3 + }} + F{e^{3 + }} + 2C{O_2} + 7{H_2}O$$
Hence the total no. of electrons involved in the reaction $$= 3$$

$$(p)$$  At anode : oxidation takes place.
$$(q)$$  At cathode : reduction takes place.
$$(r)$$  Salt bridge for migration of ions.
$$(s)$$  Electrons flow from anode to cathode.
$$(t)$$  Current flows from cathode to anode.

$${K_3}\left[ {Fe{{\left( {OH} \right)}_6}} \right] \to + 3 + x - 6 = $$       $$0 \Rightarrow x = + 3$$
$${K_2}\left[ {Fe{O_4}} \right] \to + 2 + x - 8 = 0$$       $$ \Rightarrow x = + 6$$
$$FeS{O_4} \cdot {\left( {N{H_4}} \right)_2}S{O_4} \cdot 6{H_2}O,$$       $${\text{in}}\,FeS{O_4} \to x - 2 = 0 \Rightarrow x = + 2$$
$$F{e_3}{O_4} = FeO + F{e_2}{O_3} \to $$       $$x - 2 + 2x - 6 = 0$$
$$ \Rightarrow 3x = + 8 \Rightarrow x = + \frac{8}{3}$$

$$X\,MnO_4^ - + Y\,{C_2}O_4^{2 - } + Z{H^ + } \rightleftharpoons $$       $$X\,M{n^{ + 2}} + 2Y\,C{O_2} + \frac{Z}{2}{H_2}O$$
$$\eqalign{ & {\text{First half reaction}}\,MnO_4^ - \to M{n^{ + 2}}....\left( {\text{i}} \right) \cr & {\text{On balancing}} \cr & MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O...\left( {{\text{ii}}} \right) \cr & {\text{Second half reaction}} \cr & {C_2}O_4^{2 - } \to 2C{O_2}....\left( {{\text{iii}}} \right) \cr & {\text{On balancing}} \cr & {C_2}O_4^{2 - } \to 2C{O_2} + 2{e^ - }....\left( {{\text{iv}}} \right) \cr} $$
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
$$2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to $$       $$2M{n^{ + 2}} + 10C{O_2} + 8{H_2}O$$

$${K_2}\left[ {Ni{{\left( {CN} \right)}_4}} \right]: + 2 + x + 4\left( { - 1} \right)$$       $$ = 0 \Rightarrow x = + 2$$
$${K_2}\left[ {Ni{F_6}} \right]: + 2 + x + 6\left( { - 1} \right)$$      $$ = 0 \Rightarrow x = + 4$$
$$Ni{\left( {CO} \right)_4}:x + 0 = 0 \Rightarrow x = 0$$

Metals with negative reduction potentials undergo oxidation and act as good reducing agents