Electrochemistry
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.
Given:
| Ion | $\mathrm{Z}^{\mathrm{n}+}$ | $\mathrm{U}^{\mathrm{p}+}$ | $\mathrm{V}^{\mathrm{n}+}$ | $\mathrm{X}^{\mathrm{m}-}$ | $\mathrm{Y}^{\mathrm{m}-}$ |
|---|---|---|---|---|---|
| $\lambda^{0}\left(\mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\right)$ | $50.0$ | $25.0$ | $100.0$ | $80.0$ | $100.0$ |
$\lambda^{0}$ is the limiting molar conductivity of ions
The plot of molar conductivity ( $\Lambda$ ) of $\mathrm{Z}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}} v s\, \mathrm{c}^{1 / 2}$ is given below.
Explanation:
$\Lambda_{\mathrm{V}_{\mathrm{m}} \mathrm{Y}_{\mathrm{p}}}^{\circ}=250 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\Lambda_{\mathrm{V}_m \mathrm{x}_n}=440 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
It is also given that
$ \begin{aligned} & \lambda^{\circ} \mathrm{Z}^{n+}=50 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \lambda_{\mathrm{V} p+}^0=25.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \lambda_{\mathrm{V}^{n+}}^{\circ}=100.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \lambda_{\mathrm{X}^{m-}}^{\circ}=80.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \lambda_{\mathrm{Y}^{m-}}^{\circ}=100.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ \end{aligned} $
Now, $ \Lambda_{\mathrm{V}_m \mathrm{Y}_p}=m \lambda_{\mathrm{V}^{+}}^{\circ}+p \lambda^{\circ}{ }_{\mathrm{Y}}$
$ 250=25 m+100 p $
$ 10=m+4p $ ..........(1)
Also, $ \Lambda_{\mathrm{V}_m \mathrm{x}_n}=m \lambda_{\mathrm{v}^{+}}^{\circ}+n \lambda^{\circ} \mathrm{X}^{-}$
$440=100 m+80 n$
$22=5 m+4 n$ ........(2)
In the question, a graph of $(\Lambda)$ of $Z_m X_n$ Vs $C^{1 / 2}$ is given,
We know, $ \lambda_m=\lambda_m^0-A \sqrt{C} $
For electrolyte $Z_m X_n$ and from given curve
$ \begin{aligned} & \lambda_m\left(Z_m X_n\right)=\lambda_m^0\left(Z_m X_n\right)-A \sqrt{C} \\\\ & -A=\frac{336-339}{0.04-0.01}=-\frac{3}{0.03} \\\\ & \Rightarrow A=100 \\\\ & \therefore \text { For } \lambda_m=336 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \Rightarrow 336=\lambda_m^{\circ}\left(Z_m X_n\right)-100 \times 0.04 \\\\ & \lambda_m^{\circ}=336+4=340 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \end{aligned} $
So,
$ \begin{aligned} \Lambda_{Z_m x_n} & =m \lambda_{\mathrm{Z}^{+}}+n \lambda_{\mathrm{X}^{-}} \\\\ 340 & =50 m+80 n \\\\ 34 & =5 m+8 n ........(3) \end{aligned} $
Solving eqn (2) and eqn (3),
$n=\frac{12}{4}=3 $
$ \Rightarrow $ $n=3$
Substituting the value of $n$ in eqn (2), we get,
$ \begin{aligned} 22 & =5 m+4(3) \\\\ \Rightarrow 22 & =5 m+12 \\\\ \Rightarrow 5 m & =22-12=10 \\\\ \Rightarrow m & =\frac{10}{5}=2 \\\\ \Rightarrow m & =2 \end{aligned} $
Now, substituting the value of $m$ is eqn $(1)$, we get
$ \begin{aligned} 10 & =m+4 p \\\\ \Rightarrow 10 & =2+4 p \\\\ \Rightarrow 8 & =4 p \\\\ \Rightarrow p & =\frac{8}{4}=2 \\\\ \Rightarrow p & =22 \end{aligned} $
So, $m+n+p=2+3+2$
$\Rightarrow m+n+p=7 $
Hence, the required value of $m+n+p$ is 7 .
The reduction potential $\left(E^{0}\right.$, in $\left.\mathrm{V}\right)$ of $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})$ is __________.
[Given: $E_{\left(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{MnO}_{2}(\mathrm{~s})\right)}^{0}=1.68 \mathrm{~V} ; E_{\left(\mathrm{MnO}_{2}(\mathrm{~s}) / \mathrm{Mn}^{2+}(\mathrm{aq})\right)}^{0}=1.21 \mathrm{~V} ; E_{\left(\mathrm{Mn}^{2+}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})\right)}^{0}=-1.03 \mathrm{~V}$ ]
Explanation:
(1) $MnO_4^ - (aq) + 4{H^ + } + 3e\buildrel {} \over \longrightarrow Mn{O_2}(s) + 2{H_2}O;\,E^\circ = 1.68\,V$
$\Delta G{^\circ _1} = - 3F(1.68) = - 5.04F$
(2) $MnO_2^{}(s) + 4{H^ + } + 2e\buildrel {} \over \longrightarrow M{n^{2 + }} + 2{H_2}O;\,E^\circ = 1.21\,V$
$\Delta G{^\circ _2} = - 2F(1.21) = - 2.42F$
(3) $Mn_{}^{2 + }(aq) + 2e\buildrel {} \over \longrightarrow Mn(s);\,E^\circ = - 1.03\,V$
$\Delta G{^\circ _3} = - 2F( - 1.0.3) = + 2.06F$
Adding (1), (2) and (3),
$MnO_4^ - (aq) + 8{H^ + } + 7e\buildrel {} \over \longrightarrow Mn(s) + 4{H_2}O$
$\Delta G = \Delta G{^\circ _1} + \Delta G{^\circ _2} + \Delta G{^\circ _3}$
$ = ( - 5.04 - 2.42 + 2.06)F$
$ - 7F\,E^\circ = - 5.4F$
$E$$^\circ$ = $0.77 \,V$
The $E^{\circ}$ of $\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=1.6 \mathrm{~V}, \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V}$ the $E^{\circ}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is
+0.84 V
-0.84 V
-2.32 V
+1.5 V
Electrolysis of aqueous $\mathrm{Na}_2 \mathrm{SO}_4$ was carried out by passing a current of 3 ampere for 10 min . The volume of the gas (in litre) at STP at the anode of the cell is approximately
0.19
2.1
0.10
0.15
The variation of $\lambda_{\mathrm{m}}$ of acetic acid with concentration is correctly represented as
In two separate experiments, the same quantity of electricity was passed through silver and gold solutions. [Assume ' $l$ ' constant]. The amounts of Ag and Au deposited are 2.15 and 1.31 g , respectively. The valency of gold is
[atomic mass of $\mathrm{Ag}=107.9$; $\mathrm{Au}=197$ ]
1
2
3
4
Given, $E_{\mathrm{Mn}^{7+} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}, E_{\mathrm{Mn}^{4+} / \mathrm{Mn}^{2+}}^{\circ}=1.23 \mathrm{~V}$ Calculate the $E_{\mathrm{Mn}^{7+} / \mathrm{Mn}^{4+}}^{\circ}$.
0.28 V
-0.28 V
1.7 V
0.48 V
On passing a current of 1.2 A through a solution of salt of copper for $40 \mathrm{~min}, 0.96 \mathrm{~g}$ of copper was deposited. The equivalent weight of copper in g is
21.2
31.75
63.5
15.9
$1 \mathrm{~A} / \mathrm{s}$
193000 coulombs
$ \frac{96500}{\text { (Atomic weight of the substance) }} $
Charge on 1 mole of electrons
96.5 amperes current is passed through the molten $\mathrm{AlCl}_3$ for 100 seconds. The mass of aluminium deposited at the cathode is (atomic weight of $\mathrm{Al}=27 \mathrm{u}$)
38.6 amperes of current is passed for 100 seconds through an aqueous $\mathrm{CuSO}_4$ solution using platinum electrodes. The mass of copper consumed from the solution and volume of gas liberated at STP are respectively (molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$).
The reduction potential of hydrogen electrode at $25^{\circ} \mathrm{C}$ in a neutral solution is ($p_{\mathrm{H}_2}=1$ bar)
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 :