101.
Standard electrode potential for $$\frac{{S{n^{4 + }}}}{{S{n^{2 + }}}}$$ couple is $$ + 0.15\,V$$ and that for the $$\frac{{C{r^{3 + }}}}{{Cr}}$$ couple is $$ - 0.74.$$ These two couples in their standard state are connected to make a cell. The cell potential will be A.
$$ + 0.89\,V$$
B.
$$ + 0.18\,V$$
C.
$$ + 1.83\,V$$
D.
$$ + 1.199\,V$$
Answer :
$$ + 0.89\,V$$
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$$\eqalign{
& E_{\frac{{S{n^{4 + }}}}{{S{n^{2 + }}}}}^ \circ = + 0.15\,V \cr
& E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V \cr
& E_{cell}^ \circ = E_{{\text{cathode}}\left( {RP} \right)}^ \circ - E_{{\text{anode(}}RP{\text{)}}}^ \circ \cr
& \,\,\,\,\,\,\,\,\,\,\, = 0.15 - \left( { - 0.74} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\, = + 0.89\,V \cr} $$
102.
Same amount of electric current is passed through the solutions of $$AgN{O_3}$$ and $$HCl.$$ If $$1.08\,g$$ of silver is obtained from $$AgN{O_3}$$ solution, the amount of hydrogen liberated at $$STP$$ will be A.
1.008 $$g$$
B.
11.2 $$g$$
C.
0.01 $$g$$
D.
1.1 $$g$$
Answer :
0.01 $$g$$
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$$\frac{{{W_{{H_2}}}}}{{{W_{Ag}}}} = \frac{{{E_{{H_2}}}}}{{{E_{Ag}}}}$$ (from Faraday's second law)
103.
A smuggler could not carry gold by depositing iron on the gold surface since A.
gold is denser
B.
iron rusts
C.
gold has higher reduction potential than iron
D.
gold has lower reduction potential than iron
Answer :
gold has higher reduction potential than iron
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Gold having higher $$E_{{\text{Red}}}^ \circ $$ and oxidises $$Fe \to F{e^{2 + }}.$$
104.
Limiting molar conductivity of $$N{H_4}OH$$ $$\left( {{\text{i}}{\text{.e}}{\text{.}}\,\,{{\mathop \Lambda \limits^o }_{m\,\,\left( {N{H_4}OH} \right)}}} \right)$$ is equal to A.
\[{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( N{{H}_{4}}Cl \right)}}+{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaCl \right)}}-{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaOH \right)}}\]
B.
\[{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaOH \right)}}+{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaCl \right)}}-{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( N{{H}_{4}}Cl \right)}}\]
C.
\[{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( N{{H}_{4}}OH \right)}}+{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( N{{H}_{4}}Cl \right)}}-{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( HCl \right)}}\]
D.
\[{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( N{{H}_{4}}Cl \right)}}+{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaOH \right)}}-{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaCl \right)}}\]
Answer :
\[{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( N{{H}_{4}}Cl \right)}}+{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaOH \right)}}-{{\overset{\circ }{\mathop{\Lambda }}\,}_{m\,\,\left( NaCl \right)}}\]
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According to Kohlrausch’s law,
105.
The highest electrical conductivity of the following aqueous solutions is of A.
$$0.1 M$$ difluoroacetic acid
B.
$$0.1 M$$ fluoroacetic acid
C.
$$0.1 M$$ chloroacetic acid
D.
$$0.1 M$$ acetic acid
Answer :
$$0.1 M$$ difluoroacetic acid
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Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid > fluoro acetic acid > chloro acitic acid > acetic acid.
106.
$${\text{Given}}:E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V;$$ $$E_{\frac{{MnO_4^ - }}{{M{n^{2 + }}}}}^ \circ = 1.51\,V$$ A.
$$Cl$$
B.
$${C{r^{3 + }}}$$
C.
$${M{n^{2 + }}}$$
D.
$$MnO_4^ - $$
Answer :
$$MnO_4^ - $$
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higher the value of standard reduction potential stronger will be the oxidising agent, hence $$MnO_4^ - $$ is the strongest oxidising agent.
107.
Standard electrode potentials are A.
$$F{e^{3 + }}$$ increases
B.
$$F{e^{3 + }}$$ decreases
C.
$$\frac{{F{e^{2 + }}}}{{F{e^{3 + }}}}$$ remains unchanged
D.
$$F{e^{2 + }}$$ decreases
Answer :
$$F{e^{3 + }}$$ decreases
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The metals have higher negative values of their electrode potential can displace metals having lower values from their salt solution.
108.
The standard e.m.f. of a cell involving one electron change is found to be $$0.591 V$$ at $${25^ \circ }C.$$ The equilibrium constant of the reaction is $$\left( {F = 96500\,C\,mo{l^{ - 1}};R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}} \right)$$
A.
1.0 × 1010
B.
1.0 × 105
C.
1.0 × 101
D.
1.0 × 1030
Answer :
1.0 × 1010
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$$\eqalign{
& E_{cell}^o = E_{cell}^o - \frac{{0.059}}{n}\log {K_c} \cr
& {\text{or}}\,\,0 = 0.591 - \frac{{0.0591}}{1}\log {K_c} \cr} $$
109.
$${\Delta _r}{G^ \circ }$$ for the cell with the cell reaction : $$Z{n_{\left( s \right)}} + A{g_2}{O_{\left( s \right)}} + {H_2}{O_{\left( l \right)}} \to $$ $$Zn_{\left( {aq} \right)}^{2 + } + 2A{g_{\left( s \right)}} + 2OH_{\left( {aq} \right)}^ - $$ A.
$$2.13 \times {10^5}\,J\,mo{l^{ - 1}}$$
B.
$$ - 2.13 \times {10^5}\,J\,mo{l^{ - 1}}$$
C.
$$1.06 \times {10^5}\,J\,mo{l^{ - 1}}$$
D.
$$ - 1.06 \times {10^5}\,J\,mo{l^{ - 1}}$$
Answer :
$$ - 2.13 \times {10^5}\,J\,mo{l^{ - 1}}$$
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$$\eqalign{
& E_{{\text{cell}}}^ \circ = E_{\frac{{A{g_2}O}}{{Ag}}}^ \circ - E_{\frac{{Z{n^{2 + }}}}{{Zn}}}^ \circ \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 0.344 - \left( { - 0.76} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 1.104\,V \cr
& \Delta {G^ \circ } = - nFE_{{\text{cell}}}^ \circ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = - 2 \times 96500 \times 1.104 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = - 2.13 \times {10^5}\,J\,mo{l^{ - 1}} \cr} $$
110.
The equivalent conductances of $$B{a^{2 + }}$$ and $$C{l^ - }$$ are $$127$$ and $$76\,{\Omega ^{ - 1}}\,c{m^{ - 1}}\,e{q^{ - 1}}$$ respectively at infinite dilution. The equivalent conductance of $$BaC{l_2}$$ at infinite dilution will be A.
139.52
B.
203
C.
279
D.
101.5
Answer :
139.52
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The equivalent conductance of $$BaC{l_2}$$ at infinite dilution,