Electrochemistry
Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.
For a cell, $\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \,\mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \,\mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$
the cell potential is found to be $0.43 \mathrm{~V}$ at $298 \mathrm{~K}$. The magnitude of standard electrode potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is _________ $\times 10^{-2} \mathrm{~V}$.
[Given : $E_{A{g^ + }/Ag}^\Theta $ = 0.80 V and ${{2.303RT} \over F}$ = 0.06 V]
Explanation:
Cathode : $\left.[\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})\right] 2$
Cus(s) $+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & 0.43=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left(\frac{10^{-3}}{\left(10^{-2}\right)^{2}}\right) \\ & 0.43=E_{\text {cell }}^{0}-0.03 \log 10 \\ & \mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V} \\ & \mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} \\ & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \mathrm{~V}=34 \times 10^{-2} \end{aligned} $
Resistance of a conductivity cell (cell constant $129 \mathrm{~m}^{-1}$) filled with $74.5 \,\mathrm{ppm}$ solution of $\mathrm{KCl}$ is $100 \,\Omega$ (labelled as solution 1). When the same cell is filled with $\mathrm{KCl}$ solution of $149 \,\mathrm{ppm}$, the resistance is $50 \,\Omega$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $\frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3}$. The value of $x$ is __________. (Nearest integer)
Given, molar mass of $\mathrm{KCl}$ is $74.5 \mathrm{~g} \mathrm{~mol}^{-1}$.
Explanation:
$\mathrm{KCl}$ solution $1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$
$\mathrm{KCl}$ solution $2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$
$ \begin{aligned} &\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\ &\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\ &=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\ &=\frac{50}{100} \times 2 \\ &=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\ &=1000 \end{aligned} $
The amount of charge in $\mathrm{F}$ (Faraday) required to obtain one mole of iron from $\mathrm{Fe}_{3} \mathrm{O}_{4}$ is ___________. (Nearest Integer)
Explanation:
$ x=\frac{+8}{3} $
where x is oxidation state of Fe.
$ \mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O} $
Charge required $=\frac{8}{3} \times \mathrm{F}=\frac{8 \mathrm{~F}}{3} \simeq 3 \mathrm{~F}$
The spin-only magnetic moment value of M3+ ion (in gaseous state) from the pairs Cr3+ / Cr2+, Mn3+ / Mn2+, Fe3+ / Fe2+ and Co3+ / Co2+ that has negative standard electrode potential, is ____________ B.M. [Nearest integer]
Explanation:
$\operatorname{Cr}$ (III) $\Rightarrow d^{3}$
Number of unpaired electrons $=3$
$\mu=\sqrt{3(3+2)}=\sqrt{15} \simeq 4$ B.M.
The cell potential for $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{Sn}^{x+}\right| \mathrm{Sn}$ is $0.801 \mathrm{~V}$ at $298 \mathrm{~K}$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is ____________.
$\left(\right.$ Given $: \mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}^{\mathrm{o}}=-0.763 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{x+} \mid \mathrm{Sn}}^{\mathrm{o}}=+0.008 \mathrm{~V}$ and $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)$
Explanation:
$\mathrm{C}: \mathrm{Sn}^{+\mathrm{x}}+\mathrm{xe}^{-} \rightarrow \mathrm{Sn}$
$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\mathrm{Zn} \mid \mathrm{Zn}^{2+}}^{\circ}+\mathrm{E}_{\mathrm{Sn}^{+x} \mid \mathrm{Sn}}^{\circ}$
$\Rightarrow 0.763+0.008=0.771 \mathrm{~V}$
From the Nernst equation,
$\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^{\circ} \frac{-2.303 \,\mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}$
$0.801=0.771-\frac{0.06}{\mathrm{n}} \log 10^{-2}$
$0.03=\frac{0.06}{n} \times 2$
$\mathrm{n}=4$
The cell potential for the given cell at 298 K
Pt| H2 (g, 1 bar) | H+ (aq) || Cu2+ (aq) | Cu(s)
is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10$-$x M. The value of x is ___________.
(Given : $E_{C{u^{2 + }}/Cu}^\Theta $ = 0.34 V and ${{2.303\,RT} \over F}$ = 0.06 V)
Explanation:
$ \begin{aligned} &E=E_{\text {cell }}^{\circ}-\frac{0.06}{n} \log Q \\\\ &0.31=0.34-\frac{0.06}{2} \log \frac{10^{-6}}{C} \\\\ &\log \frac{10^{-6}}{C}=1 \\\\ &C=10^{-7} M \\\\ &x=7 \end{aligned} $
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced a STP is _____________ cm3. (Nearest integer)
[Given : Faraday constant F = 96500 C mol$-$1 at STP, molar volume of an ideal gas is 22.7 L mol$-$1]
Explanation:
$0.10 \times 2 \times 3600$ coulomb produces
$ \begin{aligned} & =\frac{\frac{3}{2} \times 0.1 \times 2 \times 3600}{2 \times 96500} \\\\ & =0.0056 \text { moles of gas } \end{aligned} $
Volume of gas produced $=0.0056 \times 22.7 \mathrm{~L}$
$ \begin{aligned} & \simeq 0.127 \mathrm{~L} \\\\ & =127 \mathrm{~mL} \end{aligned} $
For the given reactions
Sn2+ + 2e$-$ $\to$ Sn
Sn4+ + 4e$-$ $\to$ Sn
the electrode potentials are ; $E_{S{n^{2 + }}/Sn}^o = - 0.140$ V and $E_{S{n^{4 + }}/Sn}^o = + 0.010$ V. The magnitude of standard electrode potential for $S{n^{4 + }}/S{n^{2 + }}$ i.e. $E_{S{n^{4 + }}/S{n^{2 + }}}^o$ is _____________ $\times$ 10$-$2 V. (Nearest integer)
Explanation:
$ \mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Sn} \quad \mathrm{E}_{2}^{0}=0.010 \mathrm{~V} $
$ \begin{aligned} & \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} \quad \mathrm{E}_{\text {cell }}^{0} \\\\ & \mathrm{E}_{\mathrm{cell}}^{\mathrm{O}}=\frac{\mathrm{n}_{2} \mathrm{E}_{2}^{\mathrm{o}}+\mathrm{n}_{1} \mathrm{E}_{1}^{0}}{\mathrm{n}}=\frac{4(0.010)+2(0.140)}{2} \\\\ & \mathrm{E}_{\text {cell }}^{0}=0.16 \mathrm{~V}=16 \times 10^{-2} \mathrm{~V} \end{aligned} $
The quantity of electricity in Faraday needed to reduce 1 mol of Cr2O$_7^{2 - }$ to Cr3+ is ____________.
Explanation:
$\because$ Each $\mathrm{Cr}$ is converting from $+6$ to $+3$
$\therefore 6$ faradays of charge is required
For the reaction taking place in the cell :
Pt (s)| H2 (g)|H+(aq) || Ag+(aq) |Ag (s)
E$_{cell}^o$ = + 0.5332 V.
The value of $\Delta$fG$^\circ$ is ______________ kJ mol$-$1. (in nearest integer)
Explanation:
# At anode, oxidation occur
H2 $\to$ 2H+ + 2e$-$ ....... (1)
# At cathode, reduction occur
2Ag+ + 2e$-$ $\to$ 2Ag ...... (2)
Adding equation (1) and (2), we get n = 2, where n = cancelled out electron
Now,
$\Delta G^\circ = - nF\,E_{cell}^o$
$ = - 2 \times 96500 \times 0.5332$
$ = - 102907.6$
$ = - 102.9$ kJ/mol
$ = - 103$ kJ/mol
The limiting molar conductivities of NaI, NaNO3 and AgNO3 are 12.7, 12.0 and 13.3 mS m2 mol$-$1, respectively (all at 25$^\circ$C). The limiting molar conductivity of AgI at this temperature is ____________ mS m2 mol$-$1.
Explanation:
(1) $\lambda_m^{\infty}(\mathrm{NaI})=12.7 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
(2) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NaNO}_3\right)=12.0 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
(3) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{AgNO}_3\right)=13.3 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
$\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=(1)+(3)-(2)$
$ =12.7+13.3-12.0 $
$ =26.0-12.0 $
$\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=14.0$
Cu(s) + Sn2+ (0.001M) $\to$ Cu2+ (0.01M) + Sn(s)
The Gibbs free energy change for the above reaction at 298 K is x $\times$ 10$-$1 kJ mol$-$1. The value of x is __________. [nearest integer]
[Given : $E_{C{u^{2 + }}/Cu}^\Theta = 0.34\,V$ ; $E_{S{n^{2 + }}/Sn}^\Theta = - 0.14\,V$ ; F = 96500 C mol$-$1]
Explanation:
A solution of Fe2(SO4)3 is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is ___________. [nearest integer]
Given : 1 F = 96500 C mol$-$1
Atomic mass of Fe = 56 g mol$-$1
Explanation:
$ \text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3} $
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3} \mathrm{~F}$
Since, charge required $=18.6 \times 10^{-3} \times 96500 \mathrm{C}$
$ =1794.9 \mathrm{C} $
And,
$ 1.5 \times \mathrm{t}=1794.9 $
$t=\frac{1794.9}{1.5 \times 60} \min$
$ \mathrm{t} \simeq 20 \mathrm{~min} $
In a cell, the following reactions take place
$\matrix{ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} & {E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.77\,V} \cr {2{I^ - } \to {I_2} + 2{e^ - }} & {E_{{I_2}/{I^ - }}^o = 0.54\,V} \cr } $
The standard electrode potential for the spontaneous reaction in the cell is x $\times$ 10$-$2 V 298 K. The value of x is ____________. (Nearest Integer)
Explanation:
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$ \begin{aligned} &=0.77-0.54 \\\\ &=0.23 \mathrm{~V} \\\\ &=23 \times 10^{-2} \mathrm{~V} \end{aligned} $
The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 $\Omega$. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 $\times$ 10$-$3 S cm$-$1, then the cell constant of the conductivity cell is ____________ $\times$ 10$-$3 cm$-$1.
Explanation:
$ \begin{array}{ll} \text { Resistance } & =1750 ~\mathrm{ohm} \\\\ \text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\ \text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \therefore \text { Cell constant } & =0.152 \times 10^{-3} \times 1750 \\\\ & =266 \times 10^{-3} \mathrm{~cm}^{-1} \end{array} $
The cell potential for the following cell
Pt |H2(g)|H+ (aq)|| Cu2+ (0.01 M)|Cu(s)
is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)
(Given : $E_{C{u^{2 + }}/Cu}^o = 0.34$ V and ${{2.303\,RT} \over F} = 0.06$ V)
Explanation:
$0.576=0.34-0.03 \log \frac{\left[\mathrm{H}^{\oplus}\right]^{2}}{[0.01]}$
$0.576-0.34=-0.03 \log \left[\mathrm{H}^{\oplus}\right]^{2}+0.03 \log (0.01)$
$ =0.06 \,\mathrm{pH}-0.06 $
$\mathrm{pH} \simeq 4.93 \simeq 5$
Match List - I with List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | $Cd(s) + 2Ni{(OH)_3}(s) \to CdO(s) + 2Ni{(OH)_2}(s) + {H_2}O(l)$ | (I) | Primary battery |
| (B) | $Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)$ | (II) | Discharging of secondary battery |
| (C) | $2PbS{O_4}(s) + 2{H_2}O(l) \to Pb(s) + Pb{O_2}(s) + 2{H_2}S{O_4}(aq)$ | (III) | Fuel cell |
| (D) | $2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l)$ | (IV) | Charging of secondary battery |
Choose the correct answer from the options given below:
Given below are two statements :
Statement I : For KI, molar conductivity increases steeply with dilution
Statement II : For carbonic acid, molar conductivity increases slowly with dilution
In the light of the above statements, choose the correct answer from the options given below :
The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is ${\Lambda _{m1}}$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is ${\Lambda _{m2}}$. The conductivities exhibited by these two cells are same. The relationship between ${\Lambda _{m2}}$ and ${\Lambda _{m1}}$ is
In which of the following half cells, electrochemical reaction is pH dependent?
In 3d series, the metal having the highest M2+/M standard electrode potential is :
The ${\left( {{{\partial E} \over {\partial T}}} \right)_P}$ of different types of half cells are as follows:
| A | B | C | D |
|---|---|---|---|
| $1 \times {10^{ - 4}}$ | $2 \times {10^{ - 4}}$ | $0.1 \times {10^{ - 4}}$ | $0.2 \times {10^{ - 4}}$ |
(Where E is the electromotive force)
Which of the above half cells would be preferred to be used as reference electrode?
The correct order of reduction potentials of the following pairs is
A. Cl2/Cl$-$
B. I2/I$-$
C. Ag+/Ag
D. Na+/Na
E. Li+/Li
Choose the correct answer from the options given below.
Explanation:
$R = {1 \over G}$ or $G = {1 \over R} = {1 \over {0.243\Omega }} = 4.115{\Omega ^{ - 1}}$
Relation between conductance (G),
conductivity ($\kappa $) and cell constant $\left( {{l \over A}} \right)$ is given as
$\kappa = {{Gl} \over A}$
$ \Rightarrow {l \over A} = {\kappa \over G} = {{1.07 \times {{10}^6}S{m^{ - 1}}} \over {4.115{\Omega ^{ - 1}}}} = 26 \times {10^4}{m^{ - 1}} \Rightarrow x = 26$
$C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \over 5}{H_2}{O_{(l)}}$ $\rightleftharpoons$ $CdS{O_4}.{9 \over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}$
The value of $E_{cell}^0$ is 4.315 V at 25$^\circ$C. If $\Delta$H$^\circ$ = $-$825.2 kJ mol$-$1, the standard entropy change $\Delta$S$^\circ$ in J K$-$1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol$-$1]
Explanation:
$ \therefore $ $\Delta$S$^\circ$ $ = {{\Delta H^\circ + nFE^\circ } \over T}$
$ = {{( - 825.2 \times {{10}^3}) + (2 \times 96487 \times 4.315)} \over {298}}$
$ = {{ - 825.2 \times {{10}^3} + 832.682 \times {{10}^3}} \over {298}}$
$ = {{7.483 \times {{10}^3}} \over {298}} = 25.11$ JK$-$1 mol$-$1
$\therefore$ Nearest integer answer is 25.
Explanation:
$ \Rightarrow { \wedge _m} = 1000 \times {{\left( {{{1.14} \over {1500}}} \right)} \over {0.001}}$ S cm2 mol$-$1
= 760 S cm2 mol$-$1
$\Rightarrow$ 760
Zn(s) + Cu2+ (0.02 M) $\to$ Zn2+ (0.04 M) + Cu(s),
Ecell = ______________ $\times$ 10$-$2 V. (Nearest integer)
[Use : $E_{Cu/C{u^{2 + }}}^0$ = $-$ 0.34 V, $E_{Zn/Z{n^{2 + }}}^0$ = + 0.76 V, ${{2.303RT} \over F} = 0.059\,V$]
Explanation:
$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$
Nernst equation = ${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$
$ \Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$
$ \Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$
$ \Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$
= 1.0911 = 109.11 $\times$ 10$-$2
= 109
(A) Sublimation enthalpy
(B) Ionisation enthalpy
(C) Hydration enthalpy
(D) Electron gain enthalpy
The total number of above properties that affect the reduction potential is ____________ (Integer answer)
Explanation:
Cu(s) | Cu2+ (aq) (0.1 M) || Ag+(aq) (0.01 M) | Ag(s)
the cell potential E1 = 0.3095 V
For the cell
Cu(s) | Cu2+ (aq) (0.01 M) || Ag+(aq) (0.001 M) | Ag(s)
the cell potential = ____________ $\times$ 10$-$2 V. (Round off the nearest integer).
[Use : ${{2.303RT} \over F}$ = 0.059]
Explanation:
$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$
Now, ${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ .... (1)
$\therefore$ ${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$ ....(2)
From (1) and (2), E2 = 0.28 V = 28 $\times$ 10$-$2 V
Explanation:
$ = 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$ S cm2 mol$-$1
$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$
HA $\rightleftharpoons$ H+ + A$-$
$0.001(1 - \alpha )0.001\alpha 0.001\alpha $
$ \Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$
$ = 12.3 \times {10^{ - 6}}$
Zn | Zn2+ (aq), (1M) || Fe3+ (aq), Fe2+ (aq) | Pt(s)
The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x $\times$ 10$-$2. The value of x is ______________. (Nearest integer)
(Given : $E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$, $E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$)
Explanation:
$2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 2{e^ - } + 2{e^{2 + }}$
$Zn + 2F{e^{3 + }}\buildrel {} \over \longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$
$E_{cell}^0 = 0.77 - (0.76)$
$ = 1.53$ V
$1.50 = 1.53 - {{0.06} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$
$\log \left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right) = {{0.03} \over {0.06}} = {1 \over 2}$
${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = {10^{1/2}} = \sqrt {10} $
${{[F{e^{3 + }}]} \over {[F{e^{2 + }}]}} = {1 \over {\sqrt {10} }}$
${{[F{e^{3 + }}]} \over {[F{e^{2 + }}] + [F{e^{3 + }}]}} = {1 \over {1 + \sqrt {10} }} = {1 \over {4.16}}$
= 0.2402
= 24 $\times$ 10-2
$C{u_{(s)}} + 2A{g^ + }(1 \times {10^{ - 3}}M) \to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$
$E_{cell}^\Theta = 2.97$ V
Ecell for the above reaction is ______________ V. (Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]
Explanation:
$ = 2.97 - {{0.059} \over 2}\log {{0.25} \over {{{({{10}^{ - 3}})}^2}}} = 2.81V$
6OH$-$ + Cl$-$ $\to$ ClO3$-$ + 3H2O + 6e$-$
A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)
(Molar mass of KClO3 = 122.6 g mol$-$1, F = 96500 C)
Explanation:
$10 = {{122.6} \over {96500 \times 6}} \times x \times 10 \times 3600$
$x = 1.311$
Ans. (1)
Explanation:
$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$
$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$
$ = 280 + 860 - 2(426)$
$ = 288$ S cm2 mol$-$1
2Fe3+(aq) + 2I$-$(aq) $ \to $ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $\Delta$rG$_m^o$ = $-$ ___________ kJ (Round off to the Nearest Integer).
$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$
Explanation:
$F{e^{3 + }} + 3{e^ - } \to Fe$
$ \therefore $ $\Delta G_1^o = - nF{E^o}$
$ = - 3F( - 0.036)$
$E_{F{e^{2 + }}/Fe}^o = 0.440V$
$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$
$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$
$ \therefore $ $\Delta G_2^o = - 2F(0.440)$
$E_{{I_2}/2{I^ - }}^o = 0.539V$
${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$
$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$
$\Delta G_3^o = - 2F( - 0.539)$
$ \therefore $ $\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$
$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$
$ = - 45934 = - 45.9KJ \simeq - 46KJ$
Explanation:
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$
= 0.58
For HCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$
$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$ Sm$-$1
Explanation:
$ \therefore $ Conductivity = 0.55 $\times$ 10$-$3 $\times$ 1.3 S cm$-$1
Molar conductivity = ${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$
$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$
= 143 S cm2 mol$-$1
= 14.3 mS m2 mol$-$1
$ \approx $ 14 mS m2 mol$-$1
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : $E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$]
Explanation:
Zn(s) + 2Ag+ $ \rightleftharpoons $ 2Ag(s) + Zn+2
$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$
$ = 0.80 - ( - 0.76)$
$ = 1.56V$
${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$
$ = 1.56 - {{0.059} \over 2} \times 3$
$ = 1.56 - 0.0885$
$ = 1.4715$
$ = 147.15 \times {10^{ - 2}}$
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^ - $ is ___________. (Integer answer)
Explanation:
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$ \therefore $ 5 moles of MnO4- required 25 F electricity.
[Given, $E_{C{u^{2 + }}/Cu}^o = 0.34$ V, $E_{NO_3^ - /NO}^o = 0.96$ V, $E_{NO_3^ - /N{O_2}}^o = 0.79$ V and at 298 K, ${{RT} \over F}$(2.303) = 0.059]
Explanation:
Cell-I $(HN{O_3} \to NO)$
$3Cu + 2NO_3^ - + 8{H^ + } \to 3C{u^{2 + }} + 2NO + 4{H_2}O$
${Q_1} = {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}}$
$\because$ $E_1^o = 0.96 - ( - 0.34) = 1.3\,V$
${E_1} = 1.3 - {{0.059} \over 6}\log {Q_1}$
Cell-II $(HN{O_3} \to N{O_2})$
$Cu + 2NO_3^ - + 4{H^ + } \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$
${Q_2} = {{[C{u^2}] \times {{({p_{N{O_2}}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^4}}}$
$\because$ $E_2^o = 0.79 - ( - 0.34)\,V = 1.13\,V$
${E_2} = 1.13 - {{0.059} \over 2}\log {Q_2}$
Now, ${E_1} = {E_2}$
$1.3 - {{0.059} \over 6}\log {Q_1} = 1.13 - {{0.059} \over 2}\log {Q_2}$
$0.17 = {{0.059} \over 6}[\log {Q_1} - 3\log {Q_2} - = {{0.059} \over 6}\log {{{Q_1}} \over {{Q_2}}}$
$ = {{0.059} \over 6}\log {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}} \times {{{{[NO_3^ - ]}^6} \times {{[{H^ + }]}^{12}}} \over {{{[C{u^{2 + }}]}^3} \times {{({p_{N{O_2}}})}^6}}}$
$ = {{0.059} \over 6}\log {{{{[{H^ + }]}^4} \times {{[NO_3^ - ]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$ [$\because$ ${p_{NO}} = {p_{N{O_2}}}$]
$ = {{0.059} \over 6}\log {{{{[HN{O_3}]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$
Now, ${p_{N{O_2}}} \equiv [HN{O_3}]$
So, $0.17 = {{0.059} \over 6}\log {[HN{O_3}]^8}$
$ = {{0.059} \over 6} \times 8\log [HN{O_3}]$
$\log [HN{O_3}] = 2.16$
$[HN{O_3}] = {10^{2.16}}M = {10^x}M$
$\therefore$ $x = 2.16$
$ \Rightarrow 2x = 2 \times 2.16 = 4.32 \simeq 4$
Explanation:
$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
n = 5
Applying Nernst equation, ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}$
or ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}$
(I) Given, [H+] = 1 M
${E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}$
(II) Now, [H+] = 10$-$4 M
${E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}$
$\therefore$ $\left| {{E_1} - {E_2}} \right|$
$\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}$
x = 3776
| List - I (Parmeter) |
List - II (Unit) |
||
|---|---|---|---|
| (a) | Cell constant | (i) | $S\,c{m^2}mo{l^{ - 1}}$ |
| (b) | Molar conductivity | (ii) | Dimensionless |
| (c) | Conductivity | (iii) | ${m^{ - 1}}$ |
| (d) | Degree of dissociation of electrolyte | (iv) | ${\Omega ^{ - 1}}{m^{ - 1}}$ |
Choose the most appropriate answer from the options given below :
Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below :
6OH- + Cl- $ \to $ ClO3- + 3H2O + 6e-
If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO3 using a current of 2 A is_________.
(Given : F = 96,500 C mol–1; molar mass of KCIO3 = 122 g mol–1)
Explanation:
$ \therefore $ To synthesise ${{10} \over {122}}$ moles of KClO3,
Charge required = ${{10} \over {122}} \times 6$ F
${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$
$ \Rightarrow $ t(hr) = ${{965 \times 100} \over {122 \times 2 \times 36}}$ = 10.98 hr $ \simeq $ 11 Hr
3 electrons are transferred has a $\Delta $Gº of 17.37 kJ mol–1 at
25 oC. The value of Eo
cell (in V) is ______ × 10–2.
(1 F = 96,500 C mol–1)
Explanation:
$\Delta $Gº = -nF$E_{cell}^o$
$ \Rightarrow $ $E_{cell}^o$ = $ - {{17.37 \times 1000} \over {3 \times 96500}}$
= -0.06 = -6.00 $ \times $ 10-2
Cr2O72- + 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
The amount of Cr3+ obtained was 0.104 g. The efficiency of the process(in%) is (Take : F = 96000 C, At. mass of chromium = 52) ______.
Explanation:
I = 2 A, t = 8 min
From 1st law of faraday,
wCr+3 = z $ \times $ i $ \times $ t
$ \Rightarrow $ wCr+3 = ${{52} \over {96000 \times 3}} \times 2 \times 8 \times 60$
= ${{52} \over {300}}$
Mass of Cr3+ ions actually obtained = 0.104 gm
% efficiency =
= ${{0.104} \over {{{52} \over {300}}}}$ $ \times $ 100
= 60 %
Pt(s) | H2 (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w0 = 2.25 eV), all other conditions remaining the same, is _______ $ \times $ 10-2 (to the nearest integer).
Given, 2.303${{RT} \over F}$ = 0.06 V;
$E_{AgCl|Ag|C{l^ - }}^0$ = 0.22 V
Explanation:
From Nernst Equation,
Ecell = $E_{cell}^0$ - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
= 0.22 – 0.06 log 10–2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 – 2.25 = 0.39 eV
$ \therefore $ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
As $\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$
0.39 = 0.22 - ${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$
$ \Rightarrow $ 0.39 = 0.22 - $0.12 \times \log \left[ {{H^ + }} \right]$
$ \Rightarrow $ 0.39 = 0.22 + 0.12 $ \times $ pH
$ \Rightarrow $ pH = 1.42 = 142 $ \times $ 10-2