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 $
$\mathrm{pH} \simeq 4.93 \simeq 5$
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
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
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$
$E_{C{u^ + }/Cu}^0 = 0.52V$
${{RT} \over F} = 0.025$)
Explanation:
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$E_{cell}^0$ = ${{RT} \over {nF}}$ln K
$ \Rightarrow $ ln K = ${{E_{cell}^0 \times nF} \over {RT}}$
= ${{0.36 \times 1} \over {0.025}}$ = 14.4 = 144 $ \times $ 10-1
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $ \to $ Cu2+(aq.) + Sn(s);
($E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$,
$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$)
Take F = 96500 C mol–1)
Explanation:
= –2 × 96500 [(–0.16) – 0.34] + RT$\left[ {{1 \over 1}} \right]$
= 96500 J
Explanation:
Moles of Ag deposited = ${{108} \over {108}}$ = 1 mole
Anode : 2H2O $ \to $ O2 + 4H+ + 4e-
Here we have to find volume of O2 evolved.
Equivalance of Ag = Equivalance of O2
$ \Rightarrow $ 1 $ \times $ 1 = nO2 $ \times $ 4
$ \Rightarrow $ nO2 = ${1 \over 4}$ mol
$ \therefore $ Volume of O2 evolved
= ${1 \over 4}$ $ \times $ 22.4
= 5.6 lit
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ when this cell attains equilibrium is _________.
(Given $E_{S{n^{2 + }}|Sn}^0 = - 0.14V$,
$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$, ${{2.303RT} \over F} = 0.06$)
Explanation:
Sn(s) + Pb+2(aq) $ \to $ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $E_{cell}^0$ - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ ....(1)
$ \Rightarrow $ $E_{cell}^0$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$
$ \Rightarrow $ $\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ = ${1 \over 3}$
$ \Rightarrow $ ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ = 2.15
2H2O $ \to $ O2 + 4H$ \oplus $ + 4e– ; $E_{red}^0$ = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
Explanation:
$ \Rightarrow $ E = 1.23 + 0.0591 × pH
$ \Rightarrow $ E = 1.23 + 0.0591 × (5)
$ \Rightarrow $ E = 1.52
Consider the above electrochemical cell where a metal electrode ( M ) is undergoing redox reaction by forming $\mathrm{M}^{+}\left(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}^{-}\right)$. The cation $\mathrm{M}^{+}$is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?
If $c_1$ is present at anode, then $c_1>c_2$.
If $c_1$ is present at cathode, then $c_1>c_2$.
If $c_1$ is present at cathode, then $c_1
If $c_1$ is present at anode, then $c_1=c_2$.
In the given electrochemical cell, $\mathrm{Ag}(\mathrm{s})|\mathrm{AgCl}(\mathrm{s})| \mathrm{FeCl}_2(\mathrm{aq}), \mathrm{FeCl}_3(\mathrm{aq}) \mid \mathrm{Pt}(\mathrm{s})$ at 298 K , the cell potential ( $\mathrm{E}_{\text {cell }}$ ) will increase when :
A. Concentration of $\mathrm{Fe}^{2+}$ is increased.
B. Concentration of $\mathrm{Fe}^{3+}$ is decreased.
C. Concentration of $\mathrm{Fe}^{2+}$ is decreased.
D. Concentration of $\mathrm{Fe}^{3+}$ is increased.
E. Concentration of $\mathrm{Cl}^{-}$is increased.
Choose the correct answer from the options given below :
C, D and E Only
A and B Only
B Only
A and E Only
Consider the following reduction processes :
$ \begin{aligned} & \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{~s}), \mathrm{E}^0=-1.66 \mathrm{~V} \\ & \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^0=+0.77 \mathrm{~V} \\ & \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^0=+1.81 \mathrm{~V} \\ & \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{~s}), \mathrm{E}^0=-0.74 \mathrm{~V} \end{aligned} $
The tendency to act as reducing agent decreases in the order :
$\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}$
$\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}$
$\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}$
$\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}$
Given below are two statements :
1 M aqueous solutions of each of Cu(NO3)2, AgNO3, Hg2(NO3)2, Mg(NO3)2 are electrolysed using inert electrodes. Given: E0Ag+/Ag = 0.80 V, E0Hg22+/Hg = 0.79 V, E0Cu2+/Cu = 0.24 V and E0Mg2+/Mg = -2.37 V.
Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.
Statement (II) : Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statements, choose the most appropriate answer from the options given below :
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively :
The standard cell potential $\left(\mathrm{E}_{\text {cell }}^{\ominus}\right)$ of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V . The standard half cell reduction potential for $\mathrm{O}_2\left(\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}\right)$ is 1.229 V .
Choose the correct statement :
$\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}$
$\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$
$\mathrm{Mg}^{2+}>\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$
$\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}$
Match List - I with List - II :
| List - I (Applications) | List - II (Batteries/Cell) |
|---|---|
| (A) Transistors | (I) Anode - Zn/Hg; Cathode - HgO + C |
| (B) Hearing aids | (II) Hydrogen fuel cell |
| (C) Inverters | (III) Anode - Zn; Cathode - Carbon |
| (D) Apollo space ship | (IV) Anode - Pb; Cathode - Pb | PbO2 |
Choose the correct answer from the options given below :
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
$\mathrm{O}_2$ gas will be evolved as a product of electrolysis of :
(A) an aqueous solution of $\mathrm{AgNO}_3$ using silver electrodes.
(B) an aqueous solution of $\mathrm{AgNO}_3$ using platinum electrodes.
(C) a dilute solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.
(D) a high concentration solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.
Choose the correct answer from the options given below :
For a Mg | Mg2+ (aq) || Ag+ (aq) | Ag the correct Nernst Equation is :