Electrochemistry-Solutions

 CBSE Class 12 Chemistry

NCERT Solutions
Chapter - 03
Electro Chemistry

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In-text Questions
1. How would you determine the standard electrode potential of the system Mg2+| Mg?
Ans. The standard electrode potential of Mg2+| Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), H2 (g)(1 atm) | H+(aq) (1M).
A cell, consisting of Mg | MgSO4(aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.
Mg | Mg2+(aq, 1M)|| H+(aq, 1M)|H2(g, 1 bar), Pt(s)
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
E⊖=E⊖R−E⊖L${\text{E}}^{\ominus }={\text{E}}_{\text{R}}^{\ominus }-{\text{E}}_{\text{L}}^{\ominus }$
Here,E⊖R${\text{E}}_{\text{R}}^{\ominus }$ for the standard hydrogen electrode is zero.
Therefore, E⊖=0−E⊖L$E^{\ominus }=0-{\text{E}}_{\text{L}}^{\ominus }$
=−E⊖L$=-{\text{E}}_{\text{L}}^{\ominus }$

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2. Can you store copper sulphate solution in a zinc pot?
Ans. Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution.
Zn + CuSO4 →$\to$ ZnSO4 + Cu
Hence, copper sulphate solution cannot be stored in a zinc pot.

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3. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Ans. Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.
Fe2+ →$\to$Fe3+ + e-1; E⊖$E^{\ominus }$= 0.77 V
This implies that the substances having higher reduction potentials than +0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are F2, Cl2, and O2.

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4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Ans. For hydrogen electrode, H+ + e-→12H2$\to \frac{1}{2}{H}_2$, it is given that pH = 10
Therefore, [H+] = 10-10 M
Now, using Nernst equation:
E = E⊖$\ominus$ - (0.0591/n) log ( 1/[H+])
E =0−0.05911log1[10−10]$0-\frac{0.0591}{1}\log \frac{1}{\left[{10}^{-10}\right]}$
= - 0.0591 log 1010 = - 0.0591(10 log 10) = - 0.591 V

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5. Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+(0.002 M) →$\to$ Ni2+(0.160 M) + 2Ag(s)
Given that E⊖(cell)=1.05V${\mathrm{E}}_{(\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l})}^{\ominus }=1.05\mathrm{V}$
Ans. Applying Nernst equation we have:
E(cell)=E⊖ (cell )−0.0591nlog[Ni2+][Ag+]2$E_{\text{ (cell })}^{\ominus }-\frac{0.0591}{n}\log \frac{\left[{\mathrm{N}\mathrm{i}}^{2+}\right]}{{\left[{\mathrm{A}\mathrm{g}}^{+}\right]}^2}$
=1.05−0.05912log(0.160)(0.002)2$=1.05-\frac{0.0591}{2}\log \frac{(0.160)}{(0.002{)}^2}$
=1.05−0.02955log0.160.000004$=1.05-0.02955\log \frac{0.16}{0.000004}$
= 1.05 - 0.02955 log 4 ×$\times$104
= 1.05 - 0.02955 (log 10000 + log 4)
= 1.05 - 0.02955 (4 + 0.6021)
= 1.05 - 0.02955 (4.6021)
= 1.05 - 0.14
= 0.91 V

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6. The cell in which the following reactions occurs:
2Fe3+(aq) + 2I-(aq) →$\to$2Fe2+(aq) + I2(s) has E⊖cell${\text{E}}_{\text{cell}}^{\ominus }$= 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Ans. Here, n = 2, E⊖cell${\text{E}}_{\text{cell}}^{\ominus }$= 0.236 V, T = 298 K
We know that:
ΔrG=−n F E⊖${\mathrm{\Delta }}_r\text{G}=-{\text{n F E}}^{\ominus }$
=−2×96487×0.236$=-2\times 96487\times 0.236$
= - 45541.864 J mol -1
= - 45.54 kJ mol -1
Again, ΔrG =−2.303RT1ogKc${\mathrm{\Delta }}_r\text{G }=-2.303\mathrm{R}\mathrm{T}1\mathrm{o}\mathrm{g}\mathrm{K}\mathrm{c}$
logKc=−ΔrG⊖2.303RT$\log {\mathrm{K}}_{\mathrm{c}}=-\frac{{\mathrm{\Delta }}_r{\mathrm{G}}^{\ominus }}{2.303\mathrm{R}\mathrm{T}}$
=−−45.54×1032.303×8.314×298$=-\frac{-45.54\times {10}^3}{2.303\times 8.314\times 298}$
= 7.981
∴$\therefore$ Kc= Antilog (7.981) = 9.57 ×$\times$ 107

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7. Why does the conductivity of a solution decrease with dilution?
Ans. The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) per unit volume decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

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8. Suggest a way to determine the A0m${\text{A}}_m^0$ value of water.
Ans. Applying Kohlrausch's law of independent migration of ions, the A0m${\text{A}}_m^0$ value of water can be determined as follows:
A0m(H2O)=λ0H+λ0OH−${\text{A}}_{m(H_{2}O)}^0={\lambda }_H^0+{\lambda }_{O{H}^{-}}^0$
=(λ0H−+λ0Cl−)+(λ0Na++λ0OH−)$=\left({\lambda }_{{\mathrm{H}}^{-}}^0+{\lambda }_{\mathrm{C}{\mathrm{l}}^{-}}^0\right)+\left({\lambda }_{\mathrm{N}{\mathrm{a}}^{+}}^0+{\lambda }_{{\mathrm{O}\mathrm{H}}^{-}}^0\right)$−(λ0Na++λ0Cl−)$-\left({\lambda }_{\mathrm{N}{\mathrm{a}}^{+}}^0+{\lambda }_{\mathrm{C}{\mathrm{l}}^{-}}^0\right)$
=A0m(HCl)+A0m(NaOH)+A0m(NaCl)$={\text{A}}_{m(HCl)}^0+{\text{A}}_{m(NaOH)}^0+{\text{A}}_{m(NaCl)}^0$
Hence, by knowing the A0m${\text{A}}_m^0$ values of HCl, NaOH, and NaCl, the A0m${\text{A}}_m^0$ value of water can be determined.

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9. The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given λ0(H+) = 349.6 S cm2 mol-1 and λ0(HCOO-) = 54.6 S cm2 mol
Ans. C = 0.025 mol L -1
λ0(H+) = 349.6 S cm2 mol-1
λ0(HCOO-) = 54.6 S cm2 mol
Λ0m(HCOOH)=λ0(H+)+λ0(HCOO−)${\mathrm{\Lambda }}_m^0(\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H})={\lambda }^0\left({\mathrm{H}}^{+}\right)+{\lambda }^0\left({\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}}^{-}\right)$
= 349.6 + 54.6
= 404.2 S cm2mol-1
Now, degree of dissociation:
α=Λm(HCOOH)Λ0m(HCOOH)$\alpha =\frac{{\mathrm{\Lambda }}_m(\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H})}{{\mathrm{\Lambda }}_m^0(\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H})}$
=46.1404.2$=\frac{46.1}{404.2}$
= 0.114(approximately)
Thus, dissociation constant:
K=c∝2(1−∝)$K=\frac{c{\propto }^2}{(1-\propto )}$
=(0.025molL−1)(0.114)2(1−0.114)$=\frac{\left(0.025\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)(0.114{)}^2}{(1-0.114)}$
=3.67 ×$\times$10-4 mol L-1

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10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Ans. I = 0.5 A
t = 2 hours = 2×60×60s$2\times 60\times 60\mathrm{s}$ = 7200 s
Thus, Q = It
= 0.5 A ×$\times$ 7200= 3600 C
We know that 96487 C = 6.023 ×$\times$ 1023 number of electrons.
Then,
3600C=6.023×1023×360096487$3600\mathrm{C}=\frac{6.023\times {10}^{23}\times 3600}{96487}$ number of electrons
= 2.25 ×$\times$1022 number of electrons
Hence, 2.25 ×$\times$1022 number of electrons will flow through the wire.

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11. Suggest a list of metals that are extracted electrolytically.
Ans. Fuel cells are voltaic cells in which the reactants are continuously supplied to the electrodes.They convert the energy from the combustion of fuels like hydrogen, CO, Methane directly into electrical energy. Methane & methanol can be used as fuels in fuel cells.

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12. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O2−7${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}$? Consider the reaction
Cr2O2−7+14H++6e−→2Cr3++8H2O${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{C}\mathrm{r}}^{3+}+8{\mathrm{H}}_2\mathrm{O}$
Ans. The given reaction is as follows:
Cr2O2−7+14H++6e−→2Cr3++8H2O${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{C}\mathrm{r}}^{3+}+8{\mathrm{H}}_2\mathrm{O}$
Therefore, to reduce 1 mole of Cr2O2−7${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}$, the required quantity of electricity will be:
= 6 F = 6×96487C$6\times 96487\text{C}$= 578922 C

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13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans. A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO2) as the cathode, and a 38% solution of sulphuric acid (H2SO4) as an electrolyte.
When the battery is in use, the following cell reactions take place:
At anode: Pb(s)+SO2−4(aq)→PbSO4)(s)+2e−${\mathrm{P}\mathrm{b}}_{(\mathrm{s})}+{\mathrm{S}\mathrm{O}}_4^{2-}(\mathrm{a}\mathrm{q})\to {\mathrm{P}\mathrm{b}\mathrm{S}\mathrm{O}}_{4)}(s)+2{\mathrm{e}}^{-}$
At cathode: PbSO4(s)+SO2−4(aq)+4H−(aq)+2e−${\mathrm{P}\mathrm{b}\mathrm{S}\mathrm{O}}_4(s)+{\mathrm{S}\mathrm{O}}_4^{2-}(\mathrm{a}\mathrm{q})+4{\mathrm{H}}^{-}(\mathrm{a}\mathrm{q})+2{\mathrm{e}}^{-}$ →PbSO4(s)+2H2O(l)$\to {\mathrm{P}\mathrm{b}\mathrm{S}\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$
The overall cell reaction is given by,
Pb(s) + PbO2(s) + 2H2SO4(aq)→$\to$2PbSO4 (s)+ 2H2O(l)
When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, PbSO4 (s) present at the anode and cathode is converted into Pb(s) and PbO2(s) respectively.

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14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Ans. Methane and methanol can be used as fuels in fuel cells.

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15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans. In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by, Fe(s) →$\to$ Fe2+(aq) + 2e-
Electrons released at the anodic spot move through the metallic object and go to another spot of the object.
There, in the presence of H+ ions, the electrons reduce oxygen. This spot behaves as the cathode. These H+ ions come either from H2CO3, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.
The reaction corresponding at the cathode is given by, O2(g) + 4H+(aq) + 4e- →$\to$2H2O(l )
The overall reaction is: 2Fe(s) + O2(g) + 4H+(aq)→$\to$ 2Fe2+(aq) + 2H2O(l)
Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide (Fe2O3.xH2O) i.e., rust.
Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.

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Chapter End Question
1. Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn
Ans. The following is the order in which the given metals displace each other from the solution of their salts. Mg, Al, Zn, Fe, Cu. Mg is highly electropositive with
Eo Mg2+/Mg = -2.37V.
Eo Al3+/Al = -1.66 V
Eo Fe2+/Fe = -0.44 V
EoZn2+/zn =-0.763V
Eo Cu2+/Cu = +0.153

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2. Given the standard electrode potentials,
K+/K = -2.93, Ag+/Ag = 0.80 V
Hg2+/Hg = 0.79 V
Mg2+/Mg = -2.73 V, Cr3+/Cr = -0.74 V
Ans. Lower the value of reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag
Hence, the reducing power of the given metals increases in the following order:
Ag < Hg < Cr < Mg < K

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3. Depict the galvanic cell in which the reaction takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans. The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(i) Zn electrode (anode) is negatively charged.
(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.
(iii) The reaction taking place at the anode is given by,
Zn(s) →$\to$Zn2+(aq) + 2e-
The reaction taking place at the cathode is given by,
2Ag+(aq) + 2e- → 2 Ag(s)

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4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd2+(aq) →$\to$2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) →$\to$Fe3+(aq) + Ag(s)
Calculate the ΔrG⊖${\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{G}}^{\ominus }$¸ and equilibrium constant of the reactions.
Ans. (i)E⊖$\ominus$ Cr3+/Cr = -0.74V
E⊖$\ominus$ Cd2+/Cd = 0.40V
The galvanic cell of the given reaction is depicted as:
Cr(s) | Cr3+(aq) || Cd2+(aq) | Cd(s)
Now, the standard cell potential is E⊖cell=E⊖R−E⊖L${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }={\mathrm{E}}_{\mathrm{R}}^{\ominus }-{\mathrm{E}}_{\mathrm{L}}^{\ominus }$
= 0.40 – (–0.74)
= +0.34 V
ΔrG⊖=−nFE⊖cell${\mathrm{\Delta }}_r{G}^{\ominus }=-nF{E}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }$
In the given equation,
n = 6
F = 96487 C mol - 1
E⊖cell$E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }$= +0.34 V
Then, ΔrG⊖=−6×96487Cmol−1×0.34V${\mathrm{\Delta }}_r{G}^{\ominus }=-6\times 96487\mathrm{C}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}\times 0.34\mathrm{V}$
= –196833.48 CV mol-1
= –196833.48 J mol-1
= –196.83 kJ mol-1
Again, ΔrG⊖=−RT ln K${\mathrm{\Delta }}_r{G}^{\ominus }=-\text{RT ln K}$
ΔtG∘=−2.303RTInK${\mathrm{\Delta }}_{\mathrm{t}}{G}^{\circ }=-2.303\mathrm{R}\mathrm{T}\mathrm{In}\mathrm{K}$
logK=−ΔtG2.303RT$\log \mathrm{K}=-\frac{{\mathrm{\Delta }}_t\mathrm{G}}{2.303\mathrm{R}\mathrm{T}}$
=196.83×1032.303×8.314×298$=\frac{196.83\times {10}^3}{2.303\times 8.314\times 298}$= 34.496
Therefore, K = antilog (34.496)
K= 3.13 ×$\times$ 1034
(ii) E⊖Fe3+/Fe2+=0.77V${\mathrm{E}}_{{\mathrm{F}\mathrm{e}}^{3+}/{\mathrm{F}\mathrm{e}}^{2+}}^{\ominus }=0.77\mathrm{V}$
E⊖Ag+/Ag=0.80V$E_{A{g}^{+}/Ag}^{\ominus }=0.80\mathrm{V}$
The galvanic cell of the given reaction is depicted as: Fe2+ (aq) | Fe3+(aq) || Ag+(aq) | Ag(s)
Now, the standard cell potential is E⊖cell=E⊖R−E⊖L${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }={\mathrm{E}}_{\mathrm{R}}^{\ominus }-{\mathrm{E}}_{\mathrm{L}}^{\ominus }$
= 0.80 – 0.77
= 0.03 V
Here, n = 1.
Then,
ΔrG⊖=−nFE⊖cell${\mathrm{\Delta }}_r{G}^{\ominus }=-nF{E}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }$
=−1×96487Cmol−1×0.03V$=-1\times 96487{\mathrm{C}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}\times 0.03\mathrm{V}$
= –2894.61 J mol-1
= –2.89 kJ mol-1
Again, ΔrG⊖=−RT ln K${\mathrm{\Delta }}_r{G}^{\ominus }=-\text{RT ln K}$
ΔrG∘=−2.303RTInK${\mathrm{\Delta }}_{\mathrm{r}}{G}^{\circ }=-2.303\mathrm{R}\mathrm{T}\mathrm{In}\mathrm{K}$
logK=−ΔrG2.303RT$\log K=-\frac{{\mathrm{\Delta }}_{r}G}{2.303\mathrm{R}\mathrm{T}}$
=−2894.612.303×8.314×298$=\frac{-2894.61}{2.303\times 8.314\times 298}$= 0.5073
Therefore, K = antilog (0.5073)
= 3.22 (approximately)

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5. Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s) I Mg2+ (0.001M ) II Cu2+ (0.0001M) I Cu(s)
(ii) Fe(s) I Fe2+ (0.001M ) II H+(1M) I H2(g)(1 bar) | Pt(s)
(iii) Sn(s) I Sn2+ (0.050M ) II H+ (0.020M) I H2(g)(1 bar) | Pt(s)
(iv) Pt(s)| Br2(l)| Br-(0.010 M) ||H+ (0.030M) I H2(g)(1 bar) | Pt(s)
Ans. (i) For the given reaction, the Nernst equation can be given as:
Ecell=E⊖cell−0.0591nlog[Mg2+][Cu2+]$E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}=E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }-\frac{0.0591}{n}\log \frac{\left[{\mathrm{M}\mathrm{g}}^{2+}\right]}{\left[{\mathrm{C}\mathrm{u}}^{2+}\right]}$
={0.34−(−2.36)}−0.05912log.001.0001$=\{0.34-(-2.36)\}-\frac{0.0591}{2}\log \frac{.001}{.0001}$
=2.7−0.05912log10$=2.7-\frac{0.0591}{2}\log 10$
= 2.7 - 0.02955 = 2.67 V (approximately)
(ii) For the given reaction, the Nernst equation can be given as:
Ecell=E⊖elll−0.0591nlog[Fe2+][H+]2$E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}=E_{\mathrm{e}\mathrm{l}\mathrm{l}\mathrm{l}}^{\ominus }-\frac{0.0591}{n}\log \frac{\left[{\mathrm{F}\mathrm{e}}^{2+}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}$
={0−(−0.44)}−0.05912log0.00112$=\{0-(-0.44)\}-\frac{0.0591}{2}\log \frac{0.001}{1^2}$
= 0.44 - 0.02955(-3)
= 0.52865 V = 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:
Ecell=E⊖cell−0.0591nlog[Sn2+][H+]2$E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}=E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }-\frac{0.0591}{n}\log \frac{\left[{\mathrm{S}\mathrm{n}}^{2+}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}$
={0−(−0.14)}−0.05912log0.050(0.020)2$=\{0-(-0.14)\}-\frac{0.0591}{2}\log \frac{0.050}{(0.020{)}^2}$
= 0.14 - 0.0295 ×$\times$ log125
= 0.14 - 0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:
Ecell=E⊖cell−0.0591nlog1[Br−]2[H+]2$E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}=E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\ominus }-\frac{0.0591}{n}\log \frac{1}{{\left[{\mathrm{B}\mathrm{r}}^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^2}$
=(0−1.09)−0.05912log1(0.010)2(0.030)2$=(0-1.09)-\frac{0.0591}{2}\log \frac{1}{(0.010{)}^2(0.030{)}^2}$
=−1.09−0.02955×log10.00000009$=-1.09-0.02955\times \log \frac{1}{0.00000009}$
=−1.09−0.02955×log19×10−8$=-1.09-0.02955\times \log \frac{1}{9\times {10}^{-8}}$
=−1.09−0.02955×log(1.11×107)$=-1.09-0.02955\times \log \left(1.11\times {10}^7\right)$
= - 1.09 - 0.02955(0.0453 + 7)
= - 1.09 - 0.208
= -1.298 V

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6. In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag2O(s) + H2O(l) →$\to$Zn2+(aq) + 2Ag(s) + 2OH-(aq). Determine ΔrG⊖${\mathrm{\Delta }}_r{G}^{\ominus }$and E⊖${\text{E}}^{\ominus }$ for the reaction.
Ans.

Therefore, E⊖${\text{E}}^{\ominus }$= 1.104 V
We know that,
ΔrG⊖=−nFE⊖${\mathrm{\Delta }}_{\mathrm{r}}{G}^{\ominus }=-{\mathrm{n}\mathrm{F}\mathrm{E}}^{\ominus }$
=−2×96487×1.04$=-2\times 96487\times 1.04$
= –213043.296 J
= –213.04 kJ

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7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans. Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol K. If ρ$\rho$ is resistivity, then we can write: K=1ρ$\mathrm{K}=\frac{1}{\rho }$
and since length is taken as 1 cm and area of cross section is 1 sq. cm so volume will be
1 cm3 so conductivity of a solution at any given concentration can also be defined as the conductance (G) of 1 cm3 volume of solution kept between two platinum electrodes.
i.e.,G=Kal$G=K\frac{a}{l}$=  k.1 = k
(Since a = 1sq. cm , l = 1 cm)
Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.
Am = KV
Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
For weak electrolyte at any concentration Am is very less compare to strong electrolyte as weak electrolyte does not dissociate completely.
The variation of Am with c√$\sqrt{c}$ for strong and weak electrolytes is shown in the following plot:

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8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.
Ans. Given,
K= 0.0248 S cm-1
c = 0.20 M
Therefore, Molar conductivity, Am=K×1000c${\mathrm{A}}_{\mathrm{m}}=\frac{\mathrm{K}\times 1000}{\mathrm{c}}$
=0.0248×10000.2$=\frac{0.0248\times 1000}{0.2}$
= 124 S cm2mol-1

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9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω$\mathrm{\Omega }$ . What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 ×$\times$10-3 S cm-1 .
Ans. Given,
Conductivity, K = 0.146 ×$\times$10-3 S cm-1
Resistance, R = 1500 Ω$\mathrm{\Omega }$
Therefore, Cell constant=K×R$=\text{K}\times \text{R}$
=0.146×10−3×1500$=0.146\times {10}^{-3}\times 1500$
= 0.219 cm-1

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Chapter End Question
10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M	0.001	0.010	0.020	0.050	0.100
102 K/S m-1	1.237	11.85	23.15	55.53	106.74

Calculate Am for all concentrations and draw a plot between Am and C0.5. Find the value of Am0.
Ans. Given, K = 1.237 ×$\times$10-2 S m-1, c = 0.001 M
Then, k = 1.237 ×$\times$ 10-2 Sm-1 ; c0.5 = 0.0316 M0.5
Therefore, Am=Kc$A_m=\frac{K}{c}$
=1.237×10−4Scm−10.001molL−1×1000cm3L$=\frac{1.237\times {10}^{-4}\mathrm{S}{\mathrm{c}\mathrm{m}}^{-1}}{0.001\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}}\times \frac{1000{\mathrm{c}\mathrm{m}}^3}{\mathrm{L}}$
= 123.7 S cm2mol-1
Given,
K = 11.85 ×$\times$10-2 S m-1, c = 0.010 M
Then, K =11.85 ×$\times$ 10-2 ; c0.5 = 0.1 M0.5
Therefore, Am=Kc$A_m=\frac{K}{c}$
=11.85×10−4Scm−10.010molL−1×1000cm3L$=\frac{11.85\times {10}^{-4}\mathrm{S}{\mathrm{c}\mathrm{m}}^{-1}}{0.010\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}}\times \frac{1000{\mathrm{c}\mathrm{m}}^3}{\mathrm{L}}$
= 118.5 S cm2mol-1
Given,
K = 23.15 ×$\times$10-4 S m-1, c = 0.020 M
Then, K = 23.15 ×$\times$10-4 S m-1, c1/2 = 0.1414 M1/2
Therefore, Am=Kc$A_m=\frac{K}{c}$
=23.15×10−4Scm−10.020molL−1×1000cm3L$=\frac{23.15\times {10}^{-4}\mathrm{S}{\mathrm{c}\mathrm{m}}^{-1}}{0.020\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}}\times \frac{1000{\mathrm{c}\mathrm{m}}^3}{\mathrm{L}}$
= 115.8 S cm2mol-1
Given,Am=Kc$A_m=\frac{K}{c}$
K = 55.53 ×$\times$10-4 S m-1, c = 0.050 M
Then, K = 55.53 ×$\times$10-4 S m-1, c1/2 = 0.2236 M1/2
Therefore,
=55.53×10−4Scm−10.050molL−1×1000cm3L$=\frac{55.53\times {10}^{-4}\mathrm{S}{\mathrm{c}\mathrm{m}}^{-1}}{0.050{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}}\times \frac{1000{\mathrm{c}\mathrm{m}}^3}{\mathrm{L}}$
= 111.11 S cm2mol-1
Given,
K = 106.74 ×$\times$10-4 S m-1, c = 0.100 M
Then, K = 106.74 ×$\times$10-4 S m-1, c1/2 = 0.3162 M 1/2
Therefore, Am = Kc$\frac{K}{c}$
=106.74×10−4Scm−10.100mol−1×1000cm3L$=\frac{106.74\times {10}^{-4}\mathrm{S}{\mathrm{c}\mathrm{m}}^{-1}}{0.100{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}\times \frac{1000{\mathrm{c}\mathrm{m}}^3}{\mathrm{L}}$
= 106.74 S cm2mol-1
Now, we have the following data:

C12M12$\frac{C^{\frac{1}{2}}}{M^{\frac{1}{2}}}$	0.0316	0.1	0.1414	0.2236	0.3162
Am(S cm2 mol-1)	123.7	118.5	115.8	111.1	106.74

Since the line interrupts Am at 124.0 S cm2mol-1, Am0 = 124.0 S cm2mol-1

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11.Conductivity of 0.00241 M acetic acid is 7.896 ×$\times$ 10 - 5 S cm - 1. Calculate its molar conductivity and if Amº for acetic acid is 390.5 S cm2 mol - 1, what is its dissociation constant?

Answer Given,K = 7.896 ×$\times$ 10 -5 S m - 1

c = 0.00241 mol L - 1
Then, molar conductivity, Am = K/c
=7.896×10−5Scm−10.00241molL−1×1000cm3L$\frac{7.896\times {10}^{-5}\mathrm{S}{\mathrm{c}\mathrm{m}}^{-1}}{0.00241\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}}\times \frac{1000{\mathrm{c}\mathrm{m}}^3}{\mathrm{L}}$
= 32.76 S cm2 mol - 1
Again, Amº = 390.5 S cm2 mol - 1
Now, α=ΛmΛ0m=32.76Scm2mol−1390.5Scm2mol−1$\alpha =\frac{{\mathrm{\Lambda }}_m}{{\mathrm{\Lambda }}_m^0}=\frac{32.76\mathrm{S}{\mathrm{c}\mathrm{m}}^2{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}{390.5\mathrm{S}{\mathrm{c}\mathrm{m}}^2{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}$
= 0.084
Dissociation constant, Ka=cα2(1−α)$K_a=\frac{c{\alpha }^2}{(1-\alpha )}$
= (0.00241 mol L-1)(0.084)2 / (1-0.084)
= 1.86 ×$\times$ 10 - 5 mol L - 1

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12. How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of MnO−4${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}$ to Mn2+.
Ans. (i) Al3+ + 3e-→$\to$Al
Therefore, Required charge = 3 F
= 3 ×$\times$ 96487 C
= 289461 C
(ii) Cu2+ + 2e-→$\to$Cu
Therefore, Required charge = 2 F
= 2 ×$\times$ 96487 C
= 192974 C
(iii) MnO−4→Mn2+${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}\to {\mathrm{M}\mathrm{n}}^{2+}$
i.e., Mn7+ + 5e-→$\to$Mn2+
Therefore, Required charge = 5 F
= 5 ×$\times$ 96487 C
= 482435 C

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13. How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2.
(ii) 40.0 g of Al from molten Al2O3.
Ans. (i) According to the question,
Ca2+ + 2e-→$\to$Ca
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium =2×2040F$\frac{2\times 20}{40}\mathrm{F}$
= 1 F
(ii) According to the question,
Al3+ + 3e-→$\to$Al
Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al = 3×4027F$\frac{3\times 40}{27}\mathrm{F}$
= 4.44 F

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14. How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2
(ii) 1 mol of FeO to Fe2O3.
Ans. (i) According to the question,
2H2O →$\to$ 4H+ + O2 + 4e-
Electricity required for the oxidation of 1 mol of H2O to O2= 2 F
= 2×$\times$ 96487 C
= 192974 C
(ii) According to the question,
Fe2+→$\to$ Fe3+ + e-
Electricity required for the oxidation of 1 mol of FeO to Fe2O3= 1 F
= 96487 C

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15. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Ans. Given,
Current = 5A
Time = 20×$\times$ 60= 1200 s
Therefore, Charge = = current × time =5×1200$=\text{ current }\times \text{ time }=5\times 1200$= 6000 C
According to the reaction,
Ni2+(aq) + 2e- →$\to$Ni(s)
Nickel deposited by 2 ×$\times$ 96487 Cof charge =58.71 g
Therefore, nickel deposited by 6000 C =58.71×60002×96487g$=\frac{58.71\times 6000}{2\times 96487}\mathrm{g}$= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.

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16. Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4 respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Ans. According to the reaction:
Ag+(aq) + e--→$\to$Ag(s)
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by = 96487×1.45108C$\frac{96487\times 1.45}{108}\mathrm{C}$= 1295.43 C
Given,
Current = 1.5 A
Therefore, Time =1295.431.5s$=\frac{1295.43}{1.5}\mathrm{s}$
= 863.6 s
= 864 s
= 14.40 min
Again,
Cu2+(aq) + 2e- →$\to$ Cu(s)
i.e., ,2 ×$\times$ 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit = 63.5×1295.432×96487g$\frac{63.5\times 1295.43}{2\times 96487}\mathrm{g}$= 0.426 g of C
Zn2+(aq) + 2e- →$\to$ Zn(s)
i.e.,2 ×$\times$ 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit =65.4×1295.432×96487g$\frac{65.4\times 1295.43}{2\times 96487}\mathrm{g}$= 0.439 g of Zn

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17. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I-(aq)
(ii) Ag+(aq) and Cu(s)
(iii) Fe3+(aq) and Br-(aq)
(iv) Ag(s) and Fe3+(aq)
(v) Br2(aq) and Fe2+(aq)
Ans.

Since E⊖$\ominus$ for the overall reaction is positive, the reaction between Fe3+(aq) and I-(aq) is feasible.

Since E⊖$\ominus$ for the overall reaction is positive, the reaction between Ag+(aq) and Cu(s) is feasible.

Since E⊖$\ominus$ for the overall reaction is negative, the reaction between Fe3+(aq) and Br-(aq) is not feasible.

Since E⊖$\ominus$ for the overall reaction is negative, the reaction between Ag(s) and Fe3+(aq) is not feasible.

Since E⊖$\ominus$ for the overall reaction is positive, the reaction between Br2(aq) and Fe2+(aq) is feasible.

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18. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Ans. (i) At cathode:
The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e-→$\to$Ag(s); E⊖$\ominus$ = 0.80 V
H+(aq) + e-→12H2$\to \frac{1}{2}{\text{H}}_2$;E⊖$\ominus$ = 0.00 V
The reaction with a higher value of E⊖${\text{E}}^{\ominus }$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by NO−3${\mathrm{N}\mathrm{O}}_3^{-}$ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.
Ag+(aq)+e−→Ag(s)${\mathrm{A}\mathrm{g}}^{+}(\mathrm{a}\mathrm{q})+{\mathrm{e}}^{-}\to {\mathrm{A}\mathrm{g}}_{(\mathrm{s})}$; E⊖=0.80V${\mathrm{E}}^{\ominus }=0.80\mathrm{V}$
H+(aq)+e−→12H2(g)${\mathrm{H}}_{(\mathrm{a}\mathrm{q})}^{+}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_{2(\mathrm{g})}$;E⊖=0.00V${\mathrm{E}}^{\ominus }=0.00\mathrm{V}$
The reaction with a higher value of E⊖${\text{E}}^{\ominus }$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode
Since Pt electrodes are inert, the anode is not attacked by NO−3${\mathrm{N}\mathrm{O}}_3^{-}$ ions. Therefore, OH- or NO−3${\mathrm{N}\mathrm{O}}_3^{-}$ions can be oxidized at the anode. But OH- ions having a lower discharge potential and get preference and decompose to liberate O2.
OH-→$\to$OH + e-
4OH- →$\to$2H2O + O2
(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.
H+(aq)+e−→12H2(g)${\mathrm{H}}_{(\mathrm{a}\mathrm{q})}^{+}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_{2(\mathrm{g})}$
At the anode, the following processes are possible.
2H2O (l) →$\to$O2(g) + 4H+ (aq) + 4e-; E⊖=+1.23V${\mathrm{E}}^{\ominus }=+1.23\mathrm{V}$.....................(1)
2SO2−4(aq)→S2O2−6(aq)+2e−$2{\mathrm{S}\mathrm{O}}_4^{2-}(\mathrm{a}\mathrm{q})\to {\mathrm{S}}_2{\mathrm{O}}_6^{2-}(\mathrm{a}\mathrm{q})+2{\mathrm{e}}^{-}$;E⊖=+1.96V${\mathrm{E}}^{\ominus }=+1.96\mathrm{V}$...................(2)
For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode: The following reduction reactions compete to take place at the cathode.
Cu2+(aq)+2e−→Cu(s)${\mathrm{C}\mathrm{u}}_{(\mathrm{a}\mathrm{q})}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{C}\mathrm{u}}_{(\mathrm{s})}$; E⊖=0.34V${\mathrm{E}}^{\ominus }=0.34\mathrm{V}$
H+(aq)+e−→1/2H2(g)${\mathrm{H}}_{(\mathrm{a}\mathrm{q})}^{+}+{\mathrm{e}}^{-}\to 1/2{\mathrm{H}}_{2(\mathrm{g})}$;E⊖=0.00V${\mathrm{E}}^{\ominus }=0.00\mathrm{V}$
The reaction with a higher value of E⊖${\text{E}}^{\ominus }$ takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
At anode: The following oxidation reactions are possible at the anode.
Cl−(aq)→12Cl2(g)+e−1${\mathrm{C}\mathrm{l}}_{(\mathrm{a}\mathrm{q})}^{-}\to \frac{1}{2}{\mathrm{C}\mathrm{l}}_{2(\mathrm{g})}+{\mathrm{e}}^{-1}$; E⊖=1.36V${\mathrm{E}}^{\ominus }=1.36\mathrm{V}$
2H2O(l)→O2(g)+4H+(zq)+4e−$2{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{l})}\to {\mathrm{O}}_{2(\mathrm{g})}+4{\mathrm{H}}_{(\mathrm{z}\mathrm{q})}^{+}+4{\mathrm{e}}^{-}$; E⊖=+1.23V${\mathrm{E}}^{\ominus }=+1.23\mathrm{V}$
At the anode, the reaction with a lower value of E⊖$E^{\ominus }$ is preferred. But due to the over-potential of oxygen, Cl- gets oxidized at the anode to produce Cl2 gas.