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71.
Arrange the following metals in increasing order of their reducing power.
$$\left[ {{\text{Given : }}E_{\frac{{{K^ + }}}{K}}^ \circ = - 2.93\,V,E_{\frac{{A{g^ + }}}{{Ag}}}^ \circ = + 0.80\,V,E_{\frac{{A{l^{3 + }}}}{{Al}}}^ \circ = - 1.66\,V,E_{\frac{{A{u^{3 + }}}}{{Au}}}^ \circ = + 1.40\,V,E_{\frac{{L{i^ + }}}{{Li}}}^ \circ = - 3.05\,V} \right]$$
A
$$Li < K < Al < Ag < Au$$
B
$$Au < Ag < Al < K < {\text{ }}Li$$
C
$$K < Al < Au < Ag < {\text{ }}Li$$
D
$$Al < Ag < Au < Li < {\text{ }}K$$
Answer :
$$Au < Ag < Al < K < {\text{ }}Li$$
Lower the electrode potential, better is the reducing power. The electrode potential increases in the order of $$\frac{{L{i^ + }}}{{Li}}\left( { - 3.05\,V} \right),\frac{{{K^ + }}}{K}\left( { - 2.93\,V} \right),$$ $$\frac{{A{l^{3 + }}}}{{Al}},\left( { - 1.66\,V} \right),\frac{{A{g^ + }}}{{Ag}}\left( { + 0.80\,V} \right)$$ and $$\frac{{A{u^{3 + }}}}{{Au}}\left( { + 1.40\,V} \right).$$ Hence, reducing power of metals will be $$Au < Ag < Al < K < {\text{ }}Li.$$
72.
In the following reaction
$$4P + 3KOH + 3{H_2}O \to 3K{H_2}P{O_2} + P{H_3}$$
A
phosphorus is both oxidised and reduced.
B
only phosphorus is reduced.
C
phosphorus is not oxidised
D
None of these
Answer :
phosphorus is both oxidised and reduced.
$$4P + 3KOH + 3{H_2}O \to 3K{H_2}P{O_2} + P{H_3}$$
$$O.N$$ of $$P = 0,$$ In $$K{H_2}P{O_2}$$ it is $$+1,$$ In $$P{H_3}$$ it is $$-3.$$
Hence $$P$$ is oxidised and reduced.
73.
Oxidation state of iron in $$Fe{\left( {CO} \right)_4}$$ is
A
+ 1
B
- 1
C
+ 2
D
0
Answer :
0
In neutral metal carbonyls, metals have zero oxidation state.
74.
How many electrons are involved in oxidation by $$KMn{O_4}$$ in basic medium?
Answer :
standard reduction potential of that species
More is $$E_{RP}^ \circ ,$$ more is the tendency to get itself reduced or more is oxidising power.
76.
Which of the following do not show disproportionation reaction?
$$ClO_4^ - ,{F_2},C{l_2},ClO_2^ - ,ClO_2^ - ,{P_4},{S_8},$$ and $$Cl{O^ - }$$
A
$$ClO_2^ - ,ClO_4^ - ,\,{\text{and}}\,Cl{O^ - }$$
B
$${F_2}\,{\text{only}}$$
C
$${F_2}\,{\text{and}}\,ClO_4^ - $$
D
$$ClO_4^ - \,{\text{only}}$$
Answer :
$${F_2}\,{\text{and}}\,ClO_4^ - $$
$${F_2}$$ being most electronegative element cannot exhibit any positive oxidation state. In $$ClO_4^ - $$ chlorine is present in its highest oxidation state i.e $$+7.$$ Therefore it does not show disproportionation reaction.
$$3B{r_2} + 6CO_3^{2 - } + 3{H_2}O \to 5B{r^ - } + BrO_3^ - + 6HCO_3^ - $$
$$O.N.$$ of $$B{r_2}$$ changes from 0 to $$–1$$ and $$+5$$ hence it is reduced as well as oxidised.