Electrochemistry
M(s) | M+ (aq ; 0.05 molar) || M+ (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.
If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be :
M(s) | M+ (aq ; 0.05 molar) || M+ (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.
For the above cell :
The value of standard electrode potential for the change,
Fe3+ (aq) + e- $\to$ Fe2+ (aq) will be
CH3OH(l) + 3/2O2 $\to$ CO2 (g) + 2H2O (l)
At 298K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2 (g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726 kJ mol−1, efficiency of the fuel cell will be
For the reduction of NO$_3^ - $ ion in an aqueous solution, E$^0$ is + 0.96 V. Values of E$^0$ for some metal ions are given below:
$\matrix{ {{V^{2 + }}(aq.) + 2{e^ - } \to V} & {{E^0} = - 1.19\,V} \cr {F{e^{3 + }}(aq.) + 3{e^ - } \to Fe} & {{E^0} = - 0.04\,V} \cr {A{u^{3 + }}(aq) + 3{e^ - } \to Au} & {{E^0} = + 1.40\,V} \cr {H{g^{2 + }}(aq) + 2{e^ - } \to Hg} & {{E^0} = + 0.86\,V} \cr } $
The pair(s) of metals that is (are) oxidized by NO$_3^ - $ in aqueous solution is(are)
Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H$_2$ gas at the cathode is (1 Faraday = 96500 C mol$^{-1}$].
$ \wedge _{C{H_3}COONa}^o$ = 91.0 S cm2/equiv
$ \wedge _{HCl}^o$ = 426.2 S cm2/equiv
What additional information/quantity one needs to calculate $ \wedge ^o$ of an aqueous solution of acetic acid?
Among the following, identify the correct statement.
While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because
Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $\mathrm{H}_{2} \mathrm{SO}_{4}$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of
The total number of moles of chlorine gas evolved is :
If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is:
The total charge (coulombs) required for complete electrolysis is:
Ag + I- $\to$ AgI + e- , Eo = 0.152 V
Ag $\to$ Ag+ + e-, Eo = -0.800 V
What is the value of log Ksp for AgI? (2.303 RT/F = 0.059 V)
Explanation:
$AgBr(s)$ $\rightleftharpoons$ $A{g^ + }(aq) + B{r^ - }(aq)$ ..... [1]
$AgN{O_3}(aq) \to A{g^ + }(aq) + NO_3^ - (aq)$ ...... [2]
Suppose the solubility of AgBr in 10$-$7 M AgNO3 is s mol L$-$1. Substituting in equation (1) and (2), we get,
$\therefore$ Total $[A{g^ + }] = (s + {10^{ - 7}})M$, ${K_{sp}}(AgBr) = [A{g^ + }][B{r^ - }]$
or, $12 \times {10^{ - 14}} = (s + {10^{ - 7}})s$ or, ${s^2} + {10^{ - 7}}s = 12 \times {10^{ - 14}}$
or, ${s^2} + {10^{ - 7}}s - 12 \times {10^{ - 14}} = 0$ or, $s = 3 \times {10^{ - 7}}M$
$\therefore$ $[B{r^ - }] = 3 \times {10^{ - 7}}M = 3 \times {10^{ - 7}} \times {10^3}{m^3} = 3 \times {10^{ - 4}}{m^3}$
$[A{g^ + }] = 3 \times {10^{ - 7}} + {10^{ - 7}} = 4 \times {10^{ - 7}}M$
$ = 4 \times {10^{ - 7}} \times {10^3}{m^3} = 4 \times {10^{ - 4}}{m^3}$
$[NO_3^ - ] = {10^{ - 7}}M = {10^{ - 7}} \times {10^3}{m^3} = 1 \times {10^{ - 4}}{m^3}$
We know, $\lambda = {\kappa \over C}$ or, $\kappa = \lambda \times C$.
$\therefore$ ${\kappa _{B{r^ - }}} = 3 \times {10^{ - 4}} \times 8 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 24 \times {10^{ - 7}}S\,{m^{ - 1}}$
$\therefore$ ${\kappa _{A{g^ + }}} = 4 \times {10^{ - 4}} \times 6 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 24 \times {10^{ - 7}}S\,{m^{ - 1}}$
$\therefore$ $\kappa _{NO_3^ - }^{} = 1 \times {10^{ - 4}} \times 7 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 7 \times {10^{ - 7}}S\,{m^{ - 1}}$
$\therefore$ ${\kappa _{total}} = {\kappa _{B{r^ - }}} + {\kappa _{A{g^ + }}} + \kappa _{NO_3^ - }^{}$
$ = 24 \times {10^{ - 7}} \times 24 \times {10^{ - 7}} + 7 \times {10^{ - 7}} = 55 \times {10^{ - 7}}S\,{m^{ - 1}}$
$ \begin{array}{r} 2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Ag}(\mathrm{~s})+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7 +2 \mathrm{H}^{+} \end{array} $
Find $\ln \mathrm{K}$ of this reaction.
66.13
58.38
28.30
46.29
When ammonia is added to the solution, pH is raised to 11 . Which half-cell reaction is affected by pH and by how much?
$\mathrm{E}_{\text {oxd }}$ will increase by a factor of 0.65 from $\mathrm{E}_{\text {oxd }}^{\mathrm{o}}$
$\mathrm{E}_{\text {oxd }}$ will decrease by a factor of 0.65 from $\mathrm{E}_{\text {oxd }}^{\mathrm{o}}$
$\mathrm{E}_{\text {red }}$ will increase by a factor of 0.65 from $\mathrm{E}_{\text {red }}^{\mathrm{o}}$
$\mathrm{E}_{\text {red }}$ will decrease by a factor of 0.65 from $\mathrm{E}_{\text {red }}^{\mathrm{o}}$
Ammonia is always added in this reaction. Which of the following must be incorrect?
$\mathrm{NH}_3$ combines with $\mathrm{Ag}^{+}$to form a complex.
$\mathrm{Ag}\left(\mathrm{NH}_3\right)_2$ is a stronger oxidising reagent than $\mathrm{Ag}^{+}$.
In absence of $\mathrm{NH}_3$, silver salt of gluconic acid is formed.
$\mathrm{NH}_3$ has affected the standard reduction potential of glucose/gluconic acid electrode.
| Electrolyte: | KCl | KNO3 | HCl | NaOAc | NaCl |
|---|---|---|---|---|---|
| ${ \wedge ^\infty }(Sc{m^2}mo{l^{ - 1}}):$ |
149.9 | 145 | 426.2 | 91 | 126.5 |
Ag+ (aq) + Cl- (aq) $\leftrightharpoons$ AgCl (s)
Given:
| Species | $\Delta G_f^o$ (kJ/mol) |
|---|---|
| Ag+ (aq) | +77 |
| Cl- (aq) | -129 |
| AgCl (s) | -109 |
Write the cell representation of above reaction and calculate $E_{cell}^o$ at 298 K. Also find the solubility product if AgCl.
(b) If 6.539 $\times$ 10-2 g of metallic zinc is added to 100 ml saturated solution of AgCl. Find the value of ${\log _{10}}{{\left[ {Z{n^{2 + }}} \right]} \over {{{\left[ {A{g^ + }} \right]}^2}}}$. How many moles of Ag will be precipitated in the above reaction. Given that
Ag+ + e- $\to$ Ag; Eo = 0.80 V;
Zn2+ + 2e- $\to$ Zn; Eo = -0.76 V;
(It was given that atomic mass of Zn = 65.39)
Explanation:
Half-cell reactions are -
$A{g^ + }(aq) + e \to Ag(s)$
$Ag(s) + C{l^ - }(aq) \to AgCl(s) + e$
Cell reaction : $A{g^ + }(aq) + C{l^ - }(aq) \to AgCl(s)$
(1) The cell is : $Ag|AgCl(s)|C{l^ - }(aq)||A{g^ + }(aq)|Ag$
$A{g^ + }(aq) + C{l^ - }(aq) \to AgCl(s)$
$\therefore$ $\Delta {G^0} = \Delta G_r^0(AgCl) - \Delta G_r^0(A{g^ + }) - \Delta G_r^0(C{l^ - })$
or, $\Delta {G^0} = [ - 109 - 77 - ( - 129)]$ kJ mol$-$1
or, $\Delta {G^0} = - 57$ kJ mol$-$1
But, $\Delta {G^0} = - nF{E^0}$ or, $ - 57000 = - 1 \times 96500 \times {E^0}$
or, ${E^0} = {{57000} \over {96500}}$ or, ${E^0} = 0.59V$
The solubility equilibrium for AgCl is
$AgCl(s)$ $\rightleftharpoons$ $A{g^ + }(aq) + C{l^ - }(aq)$
For this reaction $E_{cell}^0 = - 0.59V$
$\therefore$ ${\log _{10}}{K_{sp}} = {{nF{E^0}} \over {RT}} = - {{0.59} \over {0.059}} = - 10$
(2) Amount of zinc added $ = {{6.539 \times {{10}^{ - 2}}} \over {65.39}} = {10^{ - 3}}$ mol
Therefore, the following reactions will occur :
$2A{g^ + }(aq) + 2e \to 2Ag(s)$ ;
$Zn(s) \to Z{n^{2 + }}(aq) + 2e$ ;
$2A{g^ + }(aq) + Zn(s) \to Z{n^{2 + }}(aq) + 2Ag(s)$ ;
By Nernst equation, ${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
At equilibrium, ${E_{cell}} = 0$, $\therefore$ $1.58 = {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
or, $\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = {{1.58 \times 2} \over {0.059}} = 53.47$
$\therefore$ Equilibrium constant, $K = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$\therefore$ $\log K = 53.47$ or, $K = {10^{53.47}}$
Very high value of K indicates that the reaction goes to almost completion. Solubility of $AgCl = \sqrt {{K_{sp}}} = \sqrt {{{10}^{ - 10}}} = {10^{ - 5}}$ M
$\therefore$ $[A{g^ + }] = {10^{ - 5}}M$. Hence, number of mole of Ag+ ions in 100 mL solution = 10$-$6. Since, the reaction goes to almost completion, the amount of Ag formed = 10$-$6 mol.
(A) Calculate $\Delta_r G^\circ$ of the following reaction
$A{g^ + }(aq.) + C{l^ - }(aq.) \to AgCl(s)$
Given :
$\mathrm{\Delta_r G^\circ(AgCl)\quad-109~kJ/mole}$
$\mathrm{\Delta_r G^\circ(Cl^-)\quad-129~kJ/mole}$
$\mathrm{\Delta_r G^\circ(Ag^+)\quad-77~kJ/mole}$
(i) Represent the above reaction in form of a cell.
(ii) Calculate E$^\circ$ of the cell.
(iii) Find ${\log _{10}}{K_{sp}}$ of AgCl.
(B) If $6.539\times10^{-2}$ g of metallic Zn (amu = 65.39) was added to 100 mL of saturated solution of AgCl, then calculate ${\log _{10}} = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$. Also find how many moles of Ag will be formed.
Given that :
$\mathrm{Ag^++e^-\to Ag\quad E^\circ=0.80~V}$
$\mathrm{Zn^{2+}+2e^-\to Zn\quad E^\circ=-0.76~V}$
Explanation:
$\bullet$ The standard free energy for a given chemical reaction is equal to the difference between standard free energy of product and standard free energy of reactant.
$\bullet$ The solubility product of AgCl can be calculated using its equilibrium constant, K.
$K_{sp} =\frac{1}{K}$
(A) The chemical reaction is,
$A{g^ + }(aq.) + C{l^ - }(aq.) \to AgCl(s)$
$\mathrm{\Delta_r G^\circ(AgCl)\quad-109~kJ/mole}$
$\mathrm{\Delta_r G^\circ(Cl^-)\quad-129~kJ/mole}$
$\mathrm{\Delta_r G^\circ(Ag^+)\quad-77~kJ/mole}$
(i) To represent the above chemical reaction, we need to write the half-cell reactions first.
$Ag + {1 \over 2}C{l_2} \to AgCl$ .... (i)
$Ag \to A{g^ + } + {e^ - }$ ..... (ii)
${1 \over 2}C{l_2} + {e^ - } \to C{l^ - }$ ..... (iii)
$A{g^ + } + C{l^ - } \to AgCl$ ..... (i-ii-iii)
The cell can be represented as follows based on the above cell reaction.
$Ag|A{g^ + }||AgCl||C{l^ - }||C{l_2}|Pt$
(ii) To calculate E$^\circ$ for the cell, the formula that can be used is,
$\Delta G^\circ=-nFE^\circ$ ..... (i)
Here, $\Delta G^\circ$ denotes the standard free energy and n denotes the number of electrons transferred. The F is Faraday’s constant and E$^\circ$ represents the standard cell potential.
Given,
$\mathrm{\Delta_r G^\circ(AgCl)\quad-109~kJ/mole}$
$\mathrm{\Delta_r G^\circ(Cl^-)\quad-129~kJ/mole}$
$\mathrm{\Delta_r G^\circ(Ag^+)\quad-77~kJ/mole}$
For a given chemical reaction, the standard Gibbs free energy is the difference between the sum of Gibbs free energy of products and the sum of Gibbs free energy of reactants.
$\Delta G^\circ$ = [$\Delta G^\circ$ of AgCl - ($\Delta G^\circ$ of Ag$^+$ + $\Delta G^\circ$ of Cl$^-$)] ..... (ii)
Putting the respective values in equation (ii),
$\therefore$ $\Delta G^\circ=-109-(-129+77)=-57$ kJ/mole
For given chemical reaction, $n=1$, F = 96500C and $\Delta G^\circ=-57$ kJ/mole
Substituting the respective values in equation (i), we get
$-57=-1\times96500\times E^\circ$
$\Rightarrow E^\circ=\frac{57000}{96500}=0.59$ V
Therefore, the value of standard cell potential for the given reaction is 0.59 V
(iii) The K$_{sp}$ represents the solubility product constant for a substance, in this case AgCl.
To determine the K$_{sp}$, the formula that can be used is,
$K_{sp}=\frac{1}{K}$ ..... (iii)
Here, K is the equilibrium constant for the given chemical reaction.
The equilibrium constant can be determined by using the value of $\Delta G^\circ$.
$\Delta G^\circ = - 2.303\,RT\,\log K$ .... (iv)
$\Delta G^\circ = - {{57\,kJ} \over {mole}}$
$ = - 57000$ J/mole
R = 8.314 JK$^{-1}$ mol$^{-1}$
T = 298 K
Substitute the respective values in equation (iv), we get
$-57000=-2.303\times8.314\times298\times \log K$
$\therefore \log K=\frac{57000}{2.303\times8.314\times298}$
$=9.98 \approx 10$
$\therefore$ K = 10$^{10}$
Substitute the value of K in equation (iii) to calculate the value of K$_{sp}$ of AgCl.
${K_{sp}} = {1 \over {{{10}^{10}}}} = {10^{ - 10}}$
$\therefore$ ${\log _{10}}{K_{sp}} = - 10$
Therefore, the solubility product, K$_{sp}$ of AgCl is 10$^{-10}$.
(B) Given,
$\mathrm{Ag^++e^-\to Ag\quad E^\circ=0.80~V}$
$\mathrm{Zn^{2+}+2e^-\to Zn\quad E^\circ=-0.76~V}$
Mass of Zn = 6.539 $\times$ 10$^{-2}$ g
Atomic mass unit of Zn = 65.39
Volume of AgCl = 100 ml
To find the moles of Zn added, divide the mass of metallic Zn by atomic mass unit of Zn.
Moles of Zinc added $=\frac{6.539\times10^{-2}}{65.39}=10^{-3}$ moles
From the given half reaction, overall reaction can be represented and value of E$^\circ$ can be calculated.
The moles of metallic Zn added is 10–3 moles and from the given reaction, the number of moles of Ag added will be twice the number of moles of Zn added.
Therefore, the number of moles of Ag$^+$ added is 10$^{–6}$ moles.
The value of ${{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ can be calculated using Nernst equation.
${E_{cell}} = E_{cell}^o - {{0.0591} \over n}{\log _{10}}{{[Z{n^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$ ..... (v)
At equilibrium, ${E_{cell}} = 0$
For the given reaction, $n = 2$ and $E^\circ = 1.56$ V
Substituting the respective values in equation (v),
$0 = 1.56 - {{0.0591} \over 2}{\log _{10}}{{[Z{n^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$
${\log _{10}}{{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = {{1.56 \times 2} \over {0.0591}}$
$\therefore$ ${\log _{10}}{{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = 52.8$
Hence, the value of ${{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ is 52.8.
The equilibrium constant K can be calculated as follows :
As $K = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$\therefore$ $K = {10^{52.8}}$
Since the value of equilibrium constant is very high, the reaction almost goes to completion. Therefore, almost 100% Ag precipitate out. Thus, moles of Ag formed during the reaction will be 10$^{–5}$ moles.
[$\because$ ${K_{sp}}[AgCl] = {10^{ - 10}}$]
Final Answer :
(A) (i) $Ag|A{g^ + }||AgCl||C{l^ - }||C{l_2}|Pt$
(ii) ${E^0} = 0.59$ V
(iii) ${\log _{10}}{K_{sp}} = -10$
(B) ${\log _{10}} = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = 52.8$
Moles of Ag formed $ = {10^{ - 6}}$ moles.
$E_{F{e^{3 + }}/F{e^{2 + }}}^o$ = 0.77 V;
$E_{S{n^{2 + }}/S{n}}^o$ = -0.14 V
Under standard conditions the potential for the reaction
Sn(s) + 2Fe3+(aq) $\to$ 2Fe2+(aq) + Sn2+(aq) is :
In2+ + Cu2+ $\to$ In3+ + Cu+ at 298 K
given
$E_{C{u^{2 + }}/C{u^ + }}^o$ = 0.15 V; $E_{l{n^{2 + }}/l{n^ + }}^o$ = -0.40 V; $E_{l{n^{3 + }}/l{n^ + }}^o$ = -0.42 V;
Explanation:
We know, $\Delta {G^0} = - nF{E^0}$
(1) $C{u^{2 + }} + e \to C{u^ + }$ ; $\Delta G_1^0 = - 0.15F$
(2) $I{n^{2 + }} + e \to I{n^ + }$ ; $\Delta G_2^0 = + 0.4F$
(3) $I{n^{3 + }} + 2e \to I{n^ + }$ ; $\Delta G_3^0 = + 0.84F$
Adding equation (1) and (2) and subtracting equation (3) we get, $C{u^{2 + }} + I{n^{2 + }} \to C{u^ + } + I{n^{3 + }}$
$\Delta G_{}^0 = \Delta G_1^0 + \Delta G_2^0 - \Delta G_3^0 = - 0.59F$
$\Delta G_{}^0 = - 2.303RT\log {K_{eq}}$ $\therefore$ $ - 0.59F = - 2.303RT\log {K_{eq}}$
or, $\log {K_{eq}} = {{0.59 \times 96500} \over {2.303 \times 8.314 \times 298}} \approx 10$ $\therefore$ ${K_{eq}} = {10^{10}}$
Explanation:
The two given cells are represented as :
$Zn(s)|Z{n^{2 + }}({C_1})||C{u^{2 + }}(aq)(C = ?)|Cu(s)$, ${E_{cell}} = {E_1}$
$Zn(s)|Z{n^{2 + }}({C_2})||C{u^{2 + }}(aq)(C = 0.5\,M)|Cu(s)$, ${E_{cell}} = {E_2}$
Given, E2 > E1 , E2 $-$ E1 = 0.03 and C1 = C2 [concentration of Zn2+ is the same in both the solutions]
$\therefore$ Cell reaction : $Zn(s) + C{u^{2 + }}(aq)$ $\rightleftharpoons$ $Z{n^{2 + }}(aq) + Cu(s)$
$\therefore$ ${E_{cell}} = E_{cell}^0 - {{2.303RT} \over {2F}}\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}}$
For cell 1, ${E_1} = E_{cell}^0 - {{0.06} \over 2}\log {{{C_1}} \over C}$ [Given : ${{2.303RT} \over F} = 0.06$]
For cell 2, ${E_2} = E_{cell}^0 - {{2.303RT} \over {nF}}\log {{{C_2}} \over {0.05}}$
or, ${E_2} = E_{cell}^0 - {{0.06} \over 2}\log {{{C_2}} \over {0.05}}$
$\therefore$ ${E_2} - {E_1} = \left( {E_{cell}^0 - {{0.06} \over 2}\log {{{C_2}} \over {0.05}}} \right) - \left( {E_{cell}^0 - {{0.06} \over 2}\log {{{C_1}} \over C}} \right)$
or, $0.03 = {{0.06} \over 2}\left( {\log {{{C_2}} \over C} \times {{0.5} \over {{C_1}}}} \right) = {{0.06} \over 2}\log {{0.5} \over C}$ [$\because$ C1 = C2]
or, $\log {{0.5} \over C} = {{0.03 \times 2} \over {0.06}}$ or, ${{0.5} \over C} = 10$
$\therefore$ C = 0.05 M
$\eqalign{ & Pt({H_2})|{H^ + }(aq)|Pt({H_2}) \cr & \,\,\,\,\,{p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1M\,\,\,\,\,\,\,\,\,\,\,\,{p_2} \cr} $
$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} & \longrightarrow \mathrm{Ag}; E^{\circ}=x \\\\ \mathrm{Cu}^{2+}+2 e^{-} & \longrightarrow \mathrm{Cu}{;} E^{\circ}=y \end{aligned} $
$ E^{\circ} \text { cell is } $ :
Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s)
(i) Write the cell reaction.
(ii) Calculate $\Delta H^o$ and $\Delta S^o$m for the cell reaction by assuming that these quantities remain unchanged in the range 15oC to 35oC.
(iii) Calculate the solubility of AgCl in water at 25oC
Given : The standard reduction potential of the Ag+ (aq) / Ag (s) couple is 0.80 V at 25oC
Explanation:
Given cell : $Pt|{H_2}(g)|HCl(aq)|AgCl(s)|Ag(s)$
(1) The half-cell reactions are as follows:
At anode : ${1 \over 2}{H_2}(g) \to {H^ + }(aq) + e$
At cathode : $AgCl(s) + e \to Ag(s) + C{l^ - }(aq)$
Cell reaction : ${1 \over 2}{H_2}(g) + AgCl(s) \to {H^ + }(aq) + Ag(s) + C{l^ - }(aq)$
(2) $\Delta {S^0} = nF\left( {{{d{E^0}} \over {dT}}} \right)$, where n = Number of electrons involved in the cell reaction, F = Faraday = 96500 C, dE0 = Difference of standard electrode potential at two different temperatures = (0.21 $-$ 0.23) = $-$ 0.02 V and dT = difference of two temperatures = (308 $-$ 288) K = 20 K
$\therefore$ $\Delta {S^0} = 1 \times 96500 \times \left( {{{ - 0.02} \over {20}}} \right) = - 96.5$ J K$-$1 mol$-$1
We know, $\Delta {G^0} = - nF{E^0}$
$\therefore$ $\Delta G_{15^\circ \,C}^0 = - 1 \times 96500 \times 0.23$ [$\because$ $\Delta E_{15^\circ \,C}^0 = 0.23\,V$]
= $-$ 22195 J . mol$-$1
$\therefore$ $\Delta H_{}^0 = \Delta G_{}^0 - T\Delta S_{}^0$
$ = - 22195 - 288 \times ( - 96.5) = - 49987$ J mol$-$1
(3) Given $\Delta E_{(15^\circ \,C)}^0 = 0.23\,V$ and $\Delta E_{(35^\circ \,C)}^0 = 0.21\,V$
$\therefore$ ${{\Delta {E^0}} \over {\Delta T}} = {{(0.21 - 0.23)} \over {20}} = - 0.01$
$\therefore$ $\Delta$E0 for 10$^\circ$C = $-$0.01 $\times$ 10 = $-$0.1
$\therefore$ $\Delta E_{(25^\circ \,C)}^0 = \Delta E_{(15^\circ \,C)}^0 + ( - 0.1) = 0.23 - 0.1 = 0.22\,V$
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$
$0.22\,V = E_{C{l^ - }|AgCl|Ag}^0 - E_{2{H^ + }|{H_2}}^0 = E_{C{l^ - }|AgCl|Ag}^0 - 0$
$\therefore$ $E_{C{l^ - }|AgCl|Ag}^0$ = 0.22 V and $E_{A{g^ + }|Ag}^0$ = 0.80 V (given)
$E_{C{l^ - }|AgCl|Ag}^0 = E_{A{g^ + }|Ag}^0 - {{0.059} \over 1} \times \log {K_{sp}}(AgCl)$
or, $0.22\,V = 0.80\,V - {{0.059} \over 1}\log {K_{sp}}(AgCl)$
$\therefore$ ${K_{sp}}(AgCl) = 1.47 \times {10^{ - 10}}$
and ${K_{sp}}(AgCl) = [A{g^ + }] \times [C{l^ - }]$
$\therefore$ ${[A{g^ + }]^2} = 1.47 \times {10^{ - 10}}$
or, $[A{g^ + }] = \sqrt {1.47 \times {{10}^{ - 10}}} = 1.21 \times {10^{ - 5}}$
Explanation:
Quantity of charge passed = 2 $\times$ 10$-$3 $\times$ 16 $\times$ 60 = 1.92 C
1.92 C charge $ \equiv {1 \over {96500}} \times 1.92 \equiv 1.99 \times {10^{ - 5}}$ mol of electrons
In the reaction, $C{u^{2 + }}(aq) + 2e \to Cu(s)$
2 mol of electrons = 1 mol of Cu2+ ions discharged
$\therefore$ 1.99 $\times$ 10$-$5 mol of electrons = 9.95 $\times$ 10$-$6 of Cu2+ ions discharged.
Absorbance of a solution is directly proportional to the concentration of the solution. 50% decrease of absorbance of the solution means 50% of the concentration of the solution is reduced. Hence, initial number of mol of Cu2+ ions = 2 $\times$ 9.95 $\times$ 10$-$6 = 1.99 $\times$ 10$-$5 mol.
$\therefore$ Initial concentration of Cu2+
i.e., $CuS{O_4} = {{1.99 \times {{10}^{ - 5}}} \over {250}} \times 1000\,M = 7.96 \times {10^{ - 5}}\,M$
Pt(1) | Fe3+, Fe2+ (a = 1) | Ce4+, Ce3+ (a=1) | Pt(2)
Eo (Fe3+, Fe2+) = 0.77 V; Eo (Ce4+, Ce3+) = 1.61 V
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?
Explanation:
For electrochemical cell,
$Pt(1)|F{e^{3 + }}$ , $Fe(a = 1)||C{e^{4 + }}$ , $C{e^{3 + }}(a = 1)|Pt(2)$
$E_{F{e^{3 + }}|F{e^{2 + }}}^0 = 0.77\,V$, $E_{C{e^{4 + }}|C{e^{3 + }}}^0 = 1.61\,V$
$\therefore$ $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 1.61 - 0.77 = 0.84\,V$
Since, $E_{cell}^0$ is +ve, the cell reaction will occur spontaneously.
Hence, the flow of current will be from cathode to anode (right to left), and the current will decrease with time.