Redox Reaction MCQ Questions & Answers in Physical Chemistry | Chemistry

$$SnC{l_2} + 2FeC{l_3} \to SnC{l_4} + 2FeC{l_2}$$
Oxidation number of $$Sn$$  changes from +2 to +4, it is oxidised and acts as a reducing agent.
Oxidation number of $$Fe$$  changes from +3 to +2, it is reduced and acts as an oxidising agent.

$${K_2}C{r_2}{O_7} + 3S{O_2} + 4{H_2}S{O_4} \to $$       $${K_2}S{O_4} + C{r_2}{\left( {S{O_4}} \right)_3} + 3S{O_3} + 4{H_2}O$$
$$O.N.$$  of chrominum changes from +6 to +3

$$\left[ {Fe{{\left( {{H_2}O} \right)}_5}NO} \right]S{o_4}$$     Let $$O.N.$$  of $$Fe$$  be $$x$$ then,
$$\eqalign{ & 1 \times \left( x \right) + 5 \times \left( 0 \right) + 1 \times \left( { + 1} \right) + 1\left( { - 2} \right) = 0 \cr & \therefore \,\,x = + 1 \cr} $$

$$\eqalign{ & MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O \cr & \,\,\,\,\,\,\,\,\left[ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} \right] \times 5 \cr & \overline {MnO_4^ - + 5F{e^{2 + }} + 8{H^ + } \to M{n^{2 + }}\mathop { + 5F{e^{3 + }} + 4{H_2}O}\limits_{{\text{(in acidic medium)}}} } \cr & MnO_4^ - + 4{H^ + } + 3{e^ - } \to Mn{O_2} + 2{H_2}O \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} \right] \times 3 \cr & \overline {MnO_4^ - + 4{H^ + } + 3F{e^{2 + }} \to Mn{O_2} + \mathop {3F{e^{3 + }} + 2{H_2}O}\limits_{{\text{(in neutral medium)}}} } \cr} $$
$${\text{In acidic medium,}}$$
$$\frac{{{M_1}{V_1}}}{1} = \frac{{{M_2}{V_2}}}{5} \Rightarrow \frac{{{M_1} \times 20}}{1} = $$       $$\frac{{{M_2} \times 25}}{5},{M_1} = \frac{{{M_2}}}{4}$$
$${\text{In neutral medium,}}$$
$$\frac{{{M_1}{V_1}}}{1} = \frac{{{M_2}{V_2}}}{3} \Rightarrow \frac{{{M_2} \times {V_1}}}{4} = $$       $$\frac{{{M_2} \times 25}}{3},{V_1} = 33.3\,mL$$

$$M{g_2}{P_2}{O_7}$$
$$2\left( { + 2} \right) + 2x - 14 = 0 \Rightarrow 2x$$      $$ = 10 \Rightarrow x = + 5$$

Chlorine in $$HClO$$  is in +1 oxidation state. In $$HCl{O_4}$$  chlorine is in +7 oxidation state.

\[\begin{align} & \overset{-4}{\mathop{{{N}_{2}}}}\,\,\overset{+4}{\mathop{{{H}_{4}}}}\,\xrightarrow[N]{\text{loss}\,\,\text{of}\,{{10}_{e}}^{-}}\overset{+6}{\mathop{{{N}_{2}}}}\,Y; \\ & O.N.\,\,\text{of}\,N\,\text{changes from}\,-2\,\text{to}\,+3 \\ \end{align}\]

$$14{H^ + } + C{r_2}O_7^{2 - } + 3Ni \to $$      $$2C{r^{3 + }} + 7{H_2}O + 3N{i^{2 + }}$$
$$Ni \to N{i^{2 + }}$$   (oxidation). Hence, $$Ni$$  acts as a reducing agent.

$$B{r_2}$$  oxidises $$S$$  to a higher oxidation state $$\left( {{\text{i}}{\text{.e}}{\text{.;}}\,\,\mathop {{S_2}}\limits^{ + 2} O_3^{2 - } \to \mathop S\limits^{ + 6} O_4^{2 - }} \right)$$     and $${I_2}$$  oxidises $$S$$  to a lower oxidation state $$\left( {{\text{i}}{\text{.e}}{\text{.;}}\,\,\mathop {{S_2}}\limits^{ + 2} O_3^{2 - } \to \mathop {{S_4}}\limits^{ + 2.5} O_6^{2 - }} \right).$$     Thus, $$B{r_2}$$  is stronger oxidising agent (oxidant) than $${I_2}.$$

Zinc rod dipped in blue copper sulphate solution is oxidised to $$Z{n^{2 + }}$$  and $$C{u^{2 + }}$$  are reduced to $$Cu$$  and get deposited on zinc rod.