Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry
Learn Electrochemistry MCQ questions & answers in Physical Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
131.
$${I_2}\left( s \right)\left| {{I^ - }\left( {0.1\,M} \right)} \right.$$ half cell is connected to a $${H^ + }\left( {aq} \right)\left| {{H_2}\left( {1\,bar} \right)} \right|Pt$$ half cell and $$e.m.f.$$ is found to be $$0.7714\,V.$$ if $$E_{{I_2}\left| {{I^ - }} \right.}^ \circ = 0.535\,V,$$ find the $$pH$$ of $${H^ + }\left| {{H_2}} \right.$$ half-cell.
132.
The equivalent conductivity of $$\frac{N}{{10}}$$ solution of acetic acid at $${25^ \circ }C$$ is $$14.3\,oh{m^{ - 1}}\,c{m^2}\,equi{v^{ - 1}}.$$ What will be the degree of dissociation of acetic acid?
$$\left( {{ \wedge _{\infty \,\,C{H_3}COOH}} = 390.71\,oh{m^{ - 1}}\,c{m^2}\,equi{v^{ - 1}}} \right)$$
$$2C{r^{3 + }} + 7{H_2}O \to C{r_2}O_7^{2 - } + 14{H^ + }$$
$$O.S.$$ of $$Cr$$ changes from +3 to +6 by loss of electrons. At anode oxidation takes place.
134.
The $$e.m.f.$$ of a Daniell cell at $$298\,K$$ is $${E_1}.$$
$$Zn\left| {\mathop {ZnS{O_4}}\limits_{\left( {0.01\,M} \right)} } \right|\left| {\mathop {CuS{O_4}}\limits_{\left( {1.0\,M} \right)} } \right|Cu$$
When the concentration of $${ZnS{O_4}}$$ is $$1.0\,M$$ and that of $${CuS{O_4}}$$ is $$0.01\,M,$$ the $$e.m.f.$$ changed to $${E_2}.$$ What is the relationship between $${E_1}$$ and $${E_2}?$$
135.
$${E^ \circ }$$ value of $$\frac{{N{i^{2 + }}}}{{Ni}}$$ is $$ - 0.25\,V$$ and $$\frac{{A{g^ + }}}{{Ag}}$$ is $$ + 0.80\,V.$$ If a cell is made by taking the two electrodes what is the feasibility of the reaction?
A
Since $${E^ \circ }$$ value for the cell will be positive, redox reaction is feasible.
B
Since $${E^ \circ }$$ value for the cell will be negative, redox reaction is not feasible.
C
$$Ni$$ cannot reduce $${A{g^ + }}$$ to $$Ag$$ hence reaction is not feasible.
D
$$Ag$$ can reduce $${N{i^{2 + }}}$$ to $$Ni$$ hence reaction is feasible.
Answer :
Since $${E^ \circ }$$ value for the cell will be positive, redox reaction is feasible.
136.
Which one of the following pairs of substances on reaction will not evolve $${H_2}$$ gas?
A
Iron and $${H_2}S{O_4}\left( {aq} \right)$$
B
Iron and steam
C
Copper and $$HCl\left( {aq} \right)$$
D
Sodium and ethyl alcohol
Answer :
Copper and $$HCl\left( {aq} \right)$$
Since copper is placed below hydrogen in the electrochemical series, thus copper does not give hydrogen with dilute acids. While all other pairs give hydrogen on reaction.
$$\eqalign{
& Fe + dil.\,\,{H_2}S{O_4} \to FeS{O_4} + {H_2} \uparrow \cr
& 3Fe + \mathop {4{H_2}O}\limits_{{\text{Steam}}} \to F{e_3}{O_4} + 4{H_2} \uparrow \cr
& 2Na + {C_2}{H_5}OH \to 2{C_2}{H_5}ONa + {H_2} \uparrow \cr
& Cu + dil.\,\,HCl \to {\text{No reaction}} \cr} $$
137.
$$\eqalign{
& { \wedge _{ClC{H_2}COONa}} = 224\,oh{m^{ - 1}}\,c{m^2}g\,e{q^{ - 1}}, \cr
& { \wedge _{NaCl}} = 38.2\,oh{m^{ - 1}}c{m^2}g\,e{q^{ - 1}}, \cr
& { \wedge _{HCl}} = 203\,oh{m^{ - 1}}c{m^2}g\,e{q^{ - 1}}, \cr} $$
What is the value of $${ \wedge _{ClC{H_2}COOH}}$$
138.
The Gibbs energy for the decomposition of $$A{{\text{l}}_2}{O_3}$$ at $${500^ \circ }C$$ is as follows :
$$\frac{2}{3}A{{\text{l}}_2}{O_3} \to \frac{4}{3}A{\text{l}} + {O_2},{\Delta _r}G = + 966\,kJ\,mo{l^{ - 1}}$$
The potential difference needed for electrolytic reduction of $$A{{\text{l}}_2}{O_3}$$ at $${500^ \circ }C$$ is at least
140.
$$HN{O_3}\left( {aq} \right)$$ is titrated with $$NaOH\left( {aq} \right)$$ conductometrically, graphical representation of the titration is :
A
B
C
D
Answer :
Molar conductivity of $${H^ + }$$ and $$O{H^ - }$$ are very high as compare to other $$ions.$$
Initially conductance of solution sharply decreases due to consumption of free $${H^ + }.$$ After complete neutralization further slightly increases due to presence of $$O{H^ - }.$$