Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry
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191.
How many moles of $$Pt$$ may be deposited on the cathode when $$0.80\,F$$ of electricity is passed through a $$1.0\,M$$ solution of $$P{t^{4 + }}?$$
A
0.1 $$mol$$
B
0.2 $$mol$$
C
0.4 $$mol$$
D
0.6 $$mol$$
Answer :
0.2 $$mol$$
$$P{t^{4 + }} + 4{e^ - } \to Pt$$
$$4\,moles$$ or $$4\,F$$ of electricity is required to deposit $$1\,mole$$ of $$Pt.$$
$$0.80\,F$$ of electricity will deposit $$\frac{1}{4} \times 0.80 = 0.20\,mol$$
192.
When an aqueous solution of $$AgN{O_3}$$ is electrolysed between platinum electrodes, the substances liberated at anode and cathode are
A
silver is deposited at cathode and $${O_2}$$ is liberated at anode
B
silver is deposited at cathode and $${H_2}$$ is liberated at anode
C
hydrogen is liberated at cathode and $${O_2}$$ is liberated at anode
D
silver is deposited at cathode and $$Pt$$ is dissolved in electrolyte.
Answer :
silver is deposited at cathode and $${O_2}$$ is liberated at anode
193.
$$1.0\,L$$ each of a buffer containing $$1\,mole$$ $$N{H_3}$$ and
$$1\,mole$$ of $$NH_4^ + $$ were placed in the cathodic and anodic half-cells and $$965\,C$$ of electricity was passed. If anodic and cathodic half cells reactions involve oxidation and reduction of water only as $$2{H_2}O \to 4{H^ + } + {O_2} - 4{e^ - };$$ $$2{H_2}O + 2{e^ - } \to {H_2} + 2O{H^ - }$$
Then $$pH$$ of
A
cathodic solution will increase
B
anodic solution will decrease
C
both the solutions will remain practically constant
D
both the solutions will increase
Answer :
both the solutions will remain practically constant
Due to buffer action the pH will remain practically constant.
194.
The ionic conductivity of $$B{a^{2 + }}$$ and $$C{l^ - }$$ at infinite dilution are $$127$$ and $$76\,oh{m^{ - 1}}\,c{m^2}\,e{q^{ - 1}}$$ respectively. The equivalent conductivity of $$BaC{l_2}$$ at infinity dilution $$\left( {{\text{in}}\,oh{m^{ - 1}}c{m^2}\,e{q^{ - 1}}} \right)$$ would be :
195.
A gas $$X$$ at $$1\,atm$$ is bubbled through a solution containing a mixture of $$1\,M\,{Y^ - }$$ and $$1\,M\,{Z^ - }$$ at $${25^ \circ }C.$$ If the reduction potential of $$Z > Y > X,$$ then
A
$$Y$$ will oxidise $$X$$ and not $$Z$$
B
$$Y$$ will oxidise $$Z$$ and not $$X$$
C
$$Y$$ will oxidise both $$X$$ and $$Z$$
D
$$Y$$ will reduce both $$X$$ and $$Z.$$
Answer :
$$Y$$ will oxidise $$X$$ and not $$Z$$
As reduction potential $$Y$$ is higher than $$X$$ but lower than $$Z$$ thus $$Y$$ will oxidise $$X$$ and not $$Z.$$
196.
The pressure of $${H_2}$$ required to make the potential of $${H_2}$$ -electrode zero in pure water at $$298\,K$$ is
A
$${10^{ - 12}}atm$$
B
$${10^{ - 10}}atm$$
C
$${10^{ - 4}}atm$$
D
$${10^{ - 14}}atm$$
Answer :
$${10^{ - 14}}atm$$
From the question, we have an equation
$$\eqalign{
& 2{H^ + } + 2{e^ - } \to {H_2}\left( g \right) \cr
& {\text{According to Nernst equation,}} \cr
& E = {E^ \circ } - \frac{{0.0591}}{2}{\text{log}}\frac{{{p_{{H_2}}}}}{{{{\left[ {{H^ + }} \right]}^2}}} \cr} $$
$$ = 0 - \frac{{0.0591}}{2}{\text{log}}\frac{{{p_{{H_2}}}}}{{{{\left( {{{10}^{ - 7}}} \right)}^2}}}$$ $$\left[ {\because \,\,\left[ {{H^ + }} \right] = {{10}^{ - 7}}} \right]$$
∴ For potential of $${H_2}$$ electrode to be zero $${{p_{{H_2}}}}$$ should be equal to $${\left[ {{H^ + }} \right]^2},$$ i.e. $${10^{ - 14}}atm.$$
$$\therefore \,{\text{log}}\frac{{{{10}^{ - 14}}}}{{{{\left( {{{10}^{ - 7}}} \right)}^2}}} = 0$$
197.
The given figure shows the corrosion of iron in atmosphere,
Fill in the blanks by choosing an appropriate option. At a particular spot of an object made of iron, $$\underline {\left( {\text{i}} \right)} $$ of iron to ferrous ion takes place and that spot behaves as $$\underline {\left( {{\text{ii}}} \right)} $$ . Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $${H^ + }.$$ This spot behaves as $$\underline {\left( {{\text{iii}}} \right)} \,$$ . The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust, $$\underline {\left( {{\text{iv}}} \right)} $$ and with further production of $$\underline {\left( {\text{v}} \right)} $$ ions.
(i)
(ii)
(iii)
(iv)
(v)
(a)
oxidation
anode
cathode
Fe2O3.xH2O
hydrogen
(b)
reduction
cathode
anode
Fe3O4
hydroxide
(c)
oxidation
cathode
anode
Fe2O3.xH2O
hydrogen
(d)
oxidation
anode
cathode
Fe2O3.H2O
ferrous
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(a)
No explanation is given for this question. Let's discuss the answer together.
198.
$${E^ \circ }$$ for the cell, $$Zn\left| {Z{n^{2 + }}\left( {aq} \right)} \right|\left| {C{u^{2 + }}\left( {aq} \right)} \right|Cu$$ is $$1.10 V$$ at $${25^ \circ }C.$$ The equilibrium constant for the reaction,$$Zn\left( s \right) + C{u^{2 + }}\left( {aq} \right) \rightleftharpoons $$ $$Cu\left( s \right) + Z{n^{2 + }}\left( {aq} \right)$$ is of the order
199.
Electrolysis of an aqueous solution of $$AgN{O_3}$$ with silver electrodes produces $$\underline {\left( {\text{i}} \right)} $$ at cathode while $$\underline {\left( {{\text{ii}}} \right)} $$ ions are dissolved from anode. When $$Pt$$ electrodes are used $$\underline {\left( {{\text{iii}}} \right)} $$ is produced at anode and $$\underline {\left( {{\text{iv}}} \right)} $$ at cathode.
(i)
(ii)
(iii)
(iv)
(a)
$${H_2}$$
$$NO_3^ - $$
$$O{H^ - }$$
$${H_2}$$
(b)
$$Ag$$
$${H^ + }$$
$${O_2}$$
$${H_2}$$
(c)
$$Ag$$
$$A{g^ + }$$
$${O_2}$$
$$Ag$$
(d)
$$Ag$$
$${H^ + }$$
$$A{g^ + }$$
$${O_2}$$
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(c)
No explanation is given for this question. Let's discuss the answer together.
200.
Electrode potentials $$\left( {{E^ \circ }} \right)$$ are given below :
$$\eqalign{
& \frac{{C{u^ + }}}{{Cu}} = + 0.52\,V, \cr
& \frac{{F{e^{3 + }}}}{{F{e^{2 + }}}} = + 0.77\,V, \cr
& \frac{1}{2}\frac{{{I_2}\left( s \right)}}{{{I^ - }}} = + 0.54\,V, \cr
& \frac{{A{g^ + }}}{{Ag}} = + 0.88V. \cr} $$
Based on the above potentials, strongest oxidizing agent will be :
A
$$C{u^ + }$$
B
$$F{e^{3 + }}$$
C
$$A{g^ + }$$
D
$${I_2}$$
Answer :
$$A{g^ + }$$
Higher the value of reduction potential stronger will be the oxidising hence based
on the given values $$A{g^ + }$$ will be strongest oxidizing agent.