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131.
Consider the reaction :
$${H_2}S{O_3}\left( {aq} \right) + S{n^{4 + }}\left( {aq} \right) + {H_2}O\left( l \right)$$ $$ \to S{n^{2 + }}\left( {aq} \right) + HSO_4^ - \left( {aq} \right) + 3{H^ + }\left( {aq} \right)$$
Which of the following statements is correct ?
A
$$S{n^{4 + }}$$ is the oxidizing agent because it undergoes oxidation
B
$$S{n^{4 + }}$$ is the reducing agent because it undergoes oxidation
C
$${H_2}S{O_3}$$ is the reducing agent because it undergoes oxidation
D
$${H_2}S{O_3}$$ is the reducing agent because it undergoes reduction
Answer :
$${H_2}S{O_3}$$ is the reducing agent because it undergoes oxidation
$${H_2}\mathop S\limits^{ + 4} {O_3}\left( {aq} \right) + S{n^{4 + }}\left( {aq} \right) + {H_2}O\left( l \right)$$ $$ \to S{n^{2 + }}\left( {aq} \right) + H\mathop S\limits^{ + 6} O_4^ - \left( {aq} \right) + 3{H^ + }$$
Hence $${H_2}S{O_3}$$ is the reducing agent because it undergoes oxidation.
132.
Which compound among the following has lowest oxidation number of chlorine?
A
$$HCl{O_4}$$
B
$$HCl{O_3}$$
C
$$HCl$$
D
$$HOCl$$
Answer :
$$HCl$$
$$\eqalign{
& HCl:1 + x = 0 \Rightarrow x = - 1 \cr
& HCl{O_4}:1 + x - 8 = 0 \Rightarrow x = + 7 \cr
& HCl{O_3}:1 + x - 6 = 0 \Rightarrow x = + 5 \cr
& HOCl:1 - 2 + x = 0 \Rightarrow x = + 1 \cr} $$
133.
Which of the following is not an example of redox reaction?
$$BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl$$ is not a redox reaction as it does not involve any change in oxidation number. It is an example of double decomposition reaction.
134.
If a spoon of copper metal is placed in a solution of $$FeS{O_4},$$ what will be the correct observation?
A
Copper is dissolved in $$FeS{O_4}$$ to give brown deposit.
B
No reaction takes place.
C
Iron is deposited on copper spoon.
D
Both copper and iron are precipitated.
Answer :
No reaction takes place.
Since reduction potential of copper is higher than iron it does not get oxidised and no reaction takes place.
135.
$${E^ \circ }$$ values of some redox couples are given below. On the basis of these values choose the correct option.
$${E^ \circ }$$ values : $$\frac{{B{r_2}}}{{B{r^ - }}} = + 1.90;\frac{{A{g^ + }}}{{A{g_{\left( s \right)}}}} = + 0.80;$$ $$\frac{{C{u^{2 + }}}}{{C{u_{\left( s \right)}}}} = + 0.34;\frac{{{I_{2\left( s \right)}}}}{{{I^ - }}} = + 0.54\,V$$
A
$$Cu$$ will reduce $$B{r^ - }$$
B
$$Cu$$ will reduce $$Ag$$
C
$$Cu$$ will reduce $${I^ - }$$
D
$$Cu$$ will reduce $$B{r_2}$$
Answer :
$$Cu$$ will reduce $$B{r_2}$$
More positive the value of $${E^ \circ },$$ greater is the tendency of the species to get reduced. On the basis of the given $${E^ \circ }$$ values, $$B{r_2}$$ is having highest $${E^ \circ }$$ value (+1.90) hence, $$Cu$$ will easily reduce $$B{r_2}.$$
136.
The number of moles of $$KMn{O_4}$$ reduced by one mole of $$KI$$ in alkaline medium is
A
$$\frac{1}{5}$$
B
$$2$$
C
$$\frac{3}{2}$$
D
$$4$$
Answer :
$$2$$
$$2KMn{O_4} + {H_2}O + KI \to $$ $$2Mn{O_2} + 2KOH + KI{O_3}$$
One mole of $$KI$$ reduces 2 moles of $$KMn{O_4}.$$
137.
A mole of $${N_2}{H_4}$$ loses $$10\,mol$$ of electrons to form a new compound $$Y.$$ Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in $$Y?$$ ( There is no change in the oxidation number of hydrogen. )
A
- 1
B
- 3
C
+ 3
D
+ 5
Answer :
+ 3
From the given information, we have $${N_2}{H_4} \to Y + 10{e^ - }$$
The oxidation state of $$N$$ in $${N_2}{H_4} = - 2$$ $$\left\{ {\because 2x + 4 = 0\,\,{\text{or}}\,\,x = - 2} \right\}$$
The two nitrogen atoms in $$Y$$ (product) will balance the charge of $$10\,{e^ - }.$$ Hence the oxidation state of $$N$$ will increase by + 5, i.e., from - 2 to + 3.
138.
Consider the following reaction :
$$HCHO + 2{\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + } + 3O{H^ - }$$ $$ \to 2Ag + HCO{O^ - } + 4N{H_3} + 2{H_2}O$$
Which of the following statements regarding oxidation and reduction is correct?
A
$$HCHO$$ is oxidised to $$HCO{O^ - }$$ and $${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$$ is reduced to $$Ag.$$
B
$$HCHO$$ is reduced to $$HCO{O^ - }$$ and $${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$$ is oxidised to $$Ag.$$
C
$${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$$ is reduced to $$Ag$$ while $$O{H^ - }$$ is oxidised to $$HCO{O^ - }.$$
D
$${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$$ is oxidised to $$N{H_3}$$ while $$HCHO$$ is reduced to $${H_2}O.$$
Answer :
$$HCHO$$ is oxidised to $$HCO{O^ - }$$ and $${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$$ is reduced to $$Ag.$$
139.
Which of the following substances acts as an oxidising as well as a reducing agent?
A
$$N{a_2}O$$
B
$$SnC{l_2}$$
C
$$N{a_2}{O_2}$$
D
$$NaN{O_2}$$
Answer :
$$NaN{O_2}$$
In $$N{a_2}O,SnC{l_2}$$ and $$N{a_2}{O_2}$$ central atom is either in lowest or highest oxidation state, so they can function either as an oxidising or a reducing agent but not both. However, the oxidation state of $$N$$ in $$NaN{O_2}$$ is +3 which lies between its highest (+5) and lowest (-3) values.
140.
Which of the following statements is not correct about the given reaction?
\[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\xrightarrow{\text{Oxidation}}\] $$F{e^{3 + }} + C{O_2} + NO_3^ - $$
A
$$Fe$$ is oxidised from $$F{e^{2 + }}$$ to $$F{e^{3 + }}.$$
B
Carbon is oxidised from $${C^{2 + }}$$ to $${C^{4 + }}.$$
C
$$N$$ is oxidised from $${N^{3 - }}$$ to $${N^{5 + }}.$$
D
Carbon is not oxidised.
Answer :
Carbon is not oxidised.
In $$C{N^ - },$$ oxidation of $$C$$ is + 2, and it changes to + 4 oxidation state in $$C{O_2}.$$ So $$C$$ is also oxidised.