Q: Calculate the potential of hydrogen electrode in contact with a solution whose \$pH\$ is \$10\$.

\$\bf{Hint-}\$ When an electrode is in contact with the solution of its ions in a half-cell, it has a tendency to lose or gain electrons which are known as electrode potential.\$\bf{Explanation-}\$Given, \$pH=10\$\$pH=-\log[H^+]\$\$\Rightarrow [H^+]=10^{-10}M\$Also for hydrogen electrode : \$E_{cell}=E^0-\dfrac{0.0591}{1}\log \dfrac{|H_2|}{|H^+|}\$(no of electron included \$=1\$)\$\boxed{E^o=0}\$\$E_{cell}=\dfrac{-0.0591}{1}\log\left(\dfrac{1}{10^{-10}}\right)\$\$\boxed{E_{cell}=-0.591V}\$

Q: Calculate the emf of the cell in which of the following reaction takes place:\$Ni(s) + 2Ag^+ (0.002 M) \rightarrow Ni^{2+} (0.160 M) + 2Ag(s)\$[Given that \$E^{\circ}_{cell} = 1.05 V\$]

Given,\$[Ag^+]=0.002M\$\$[Ni^{2+}]=0.160M\$\$n=2\$According to Nernst equation:\$E_{cell}=E^0-\dfrac{0.0591}{2}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2}\$\$\therefore E_{cell}=1.05-\dfrac{0.0591}{2}\log \dfrac{0.160}{(0.002)^2}\$\$\therefore E_{cell}=1.05-0.02955\, \log{(4\times 10^4)}\$\$\therefore E_{cell}=0.91V\$Hence, the correct option is \$\text{B}\$

Q: Calculate the emf of the cell. \$Zn | Zn^{2+} (0.001 M) || Cu^{2+} (0.1 M) | Cu\$The standard potential of \$Cu/Cu^{2+}\$ half-cell is +0.34 and \$Zn/Zn^{2+}\$ is -0.76 V.

Given \$Zn|Zn^{2+}(0.001M)||Cu^{2+}(0.1M)|Cu\$Overall cell reaction:\$Zn\longrightarrow { Zn }^{ 2+ }+2{ e }^{ - }\\ \underline { { Cu }^{ 2+ }+2{ e }^{ - }\longrightarrow Cu\quad \quad \quad } \\ \underline { Zn+{ Cu }^{ 2+ }\longrightarrow { Zn }^{ 2+ }+Cu } \$\$E_{cell}^{o}=\$standard reduction potential of cathode + standard oxidation potential of anode\$E_{cell}^{o}=\$0.34 to 0.76 V\$E_{cell}^{o}=\$1.1 V\${ K }_{ C }=\cfrac { \left[ { Zn }^{ 2+ } \right] }{ \left[ { Cu }^{ 2+ } \right] } =\cfrac { 10^{ -3 } }{ 10^{ -1 } } =10^{ -2 }\$EMF of the cell at any electrode concentration is:\$E=E^{ o }-\cfrac { 0.059 }{ n } \log \left( { K }_{ C } \right) =1.1-\cfrac { 0.059 }{ 2 } \log \left( 10^{ -2 } \right) =1.1-\cfrac { 0.059 }{ 2 } \times (2)=1.1-0.059=1.041V\$

Q: The conductivity of 0.20 M solution of \$KCl\$ at 298 K is \$0.0248 S cm^{-1}\$. Calculate its molar conductivity.

\$k\$ (Conductivity) \$=0.0248 \ Scm^{-1}\$\$C\$ (molarity) \$=0.20 \ M\$\$T\$ (temperature) \$=298 \ K\$\$ \Lambda _m =?\$ (molar conductivity) The formula is \$\Lambda_m = \cfrac {1000 \times k}{C}\$\$= \cfrac {1000 \times 0.0248 \ Scm^{-1}}{0.20 \ mol\ cm^{-3}}\$\$=124 \ S \ cm^2 \ mol^{-1}\$The molar conductivity is \$124 \ S \ cm^2 \ mol^{-1}\$

Q: Predict the products of electrolysis in each of the following:(i) An aqueous solution of \$AgNO_{3}\$ with silver electrodes.(ii) An aqueous solution of \$AgNO_{3}\$ with platinum electrodes.(iii) A dilute solution of \$H_{2}SO_{4}\$ with platinum electrodes.(iv) An aqueous solution of \$CuCl_{2}\$ with platinum electrodes

(i) An aqueous solution of \$AgNO_{3}\$ with silver electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Silver anode will dissolve to form silver ions in the solution.\$\displaystyle Ag \rightarrow Ag^+ + e^-\$(ii) An aqueous solution of \$AgNO_{3}\$ with platinum electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.\$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-\$(iii) A dilute solution of \$H_{2}SO_{4}\$ with platinum electrodes.At cathode: \$\displaystyle 2H^+ + 2e^- \rightarrow H_2(g)\$At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.\$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-\$(iv) An aqueous solution of \$CuCl_{2}\$ with platinum electrodesAt cathode: Cupric ions will be reduced in preference to protons\$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu\$At anode: Chloride ions will be oxidized in preference to hydroxide ions\$\displaystyle 2Cl^- \rightarrow Cl_2 + 2e^-\$

Q: Resistance and conductivity of a cell containing 0.001 M KCl solution at 298 K are \$1500 \Omega\$ and \$1.46 \times 10^{-4} S.cm^{-1}\$ respectively. What is the cell constant ?

\$R=1500 \ \Omega\$ (Resistance)\$k= 1.46 \times 10^{-4} \ Scm^{-1}\$ (Conductivity) \$T=298 \ K\$We will use the formula, \$\boxed {k=Gx}\$where \$k\$ is conductivity\$G\$ is \$\cfrac 1R\$\$x\$ is cell constant\$1.46 \times 10^{-4} = \cfrac 1{1500} \times x\$\$1.46 \times 10^{-4} \times 1500 =x\$\$\boxed {0.219 \ cm^{-1}=x}\$The cell constant is \$0.219 \ cm^{-1}\$

Q: In electrolysis of water: (a) Name the gas collected at cathode and anode.(b) Why is volume of one gas collected at one electrode is double of anode?(c) Why are few drops of dilute \$H_2SO_4\$ added to water?

a) At cathode, \$H_2\$ is evolved and at anode \$O_2\$ is evolved.b) At cathode, \$H^+\$ ion takes two electron to convert itself into \$H_2\$ gas. \$2\$ moles of \$H^+\$ gives \$1\$ mole of \$H_2\$.\$2H^+(aq)+ 2e^- \rightarrow H_2(g)\$At anode, \$OH^-\$ ion releases two electrons to convert into water and \$O_2\$. \$2\$ mole \$OH^-\$ gives \$0.5\$ mole \$O_2\$, i.e why volume of one gas collected at one electron is double of another.c) Acid is added to make the water conduct electricity as the distilled water is a non-conductor of electricity.

Q: How many electrons are there in one coulomb of electricity?

Change on \$1e^-=1.602\times 10^{-19}\$ \$C\$If total charge \$=1\$ \$C\$\$\therefore\$ Total no. of \$e^-=\cfrac{1}{1.602\times 10^{-19}}=6.24\times 10^{18}e^-\$Hence, the correct option is \$(C)\$.

Q: A metal carbonate X on reacting with an acid gives a gas which when passed through a solution gives the carbonate black. On the other hand, A gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives compound Z, used for disinfecting water. Identify X, Y, G and Z.

According to the problem:Here,\$X\$ is calcium carbonate \$\left( {CaC{O_3}} \right)\$.\$Y\$ is slaked lime.\$Z\$ is bleaching powder \$\left( {CaOC{l_2}} \right)\$\$G\$ is chlorine gas. Hence the reactions are involved,

Question 1

Calculate the potential of hydrogen electrode in contact with a solution whose $pH$ is $10$.

Accepted Answer

$\bf{Hint-}$ When an electrode is in contact with the solution of its ions in a half-cell, it has a tendency to lose or gain electrons which are known as electrode potential.$\bf{Explanation-}$Given, $pH=10$$pH=-\log[H^+]$$\Rightarrow [H^+]=10^{-10}M$Also for hydrogen electrode : $E_{cell}=E^0-\dfrac{0.0591}{1}\log \dfrac{|H_2|}{|H^+|}$(no of electron included $=1$)$\boxed{E^o=0}$$E_{cell}=\dfrac{-0.0591}{1}\log\left(\dfrac{1}{10^{-10}}\right)$$\boxed{E_{cell}=-0.591V}$

Question 2

Calculate the emf of the cell in which of the following reaction takes place:$Ni(s) + 2Ag^+ (0.002 M) \rightarrow Ni^{2+} (0.160 M) + 2Ag(s)$[Given that $E^{\circ}_{cell} = 1.05 V$]

Accepted Answer

Given,$[Ag^+]=0.002M$$[Ni^{2+}]=0.160M$$n=2$According to Nernst equation:$E_{cell}=E^0-\dfrac{0.0591}{2}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2}$$\therefore E_{cell}=1.05-\dfrac{0.0591}{2}\log \dfrac{0.160}{(0.002)^2}$$\therefore E_{cell}=1.05-0.02955\, \log{(4\times 10^4)}$$\therefore E_{cell}=0.91V$Hence, the correct option is $\text{B}$

Question 3

Calculate the emf of the cell. $Zn | Zn^{2+} (0.001 M) || Cu^{2+} (0.1 M) | Cu$The standard potential of $Cu/Cu^{2+}$ half-cell is +0.34 and $Zn/Zn^{2+}$ is -0.76 V.

Accepted Answer

Given $Zn|Zn^{2+}(0.001M)||Cu^{2+}(0.1M)|Cu$Overall cell reaction:$Zn\longrightarrow { Zn }^{ 2+ }+2{ e }^{ - }\ \underline { { Cu }^{ 2+ }+2{ e }^{ - }\longrightarrow Cu\quad \quad \quad  } \ \underline { Zn+{ Cu }^{ 2+ }\longrightarrow { Zn }^{ 2+ }+Cu } $$E_{cell}^{o}=$standard reduction potential of cathode + standard oxidation potential of anode$E_{cell}^{o}=$0.34 to 0.76 V$E_{cell}^{o}=$1.1 V${ K }_{ C }=\cfrac { \left[ { Zn }^{ 2+ } \right]  }{ \left[ { Cu }^{ 2+ } \right]  } =\cfrac { 10^{ -3 } }{ 10^{ -1 } } =10^{ -2 }$EMF of the cell at any electrode concentration is:$E=E^{ o }-\cfrac { 0.059 }{ n } \log  \left( { K }_{ C } \right) =1.1-\cfrac { 0.059 }{ 2 } \log  \left( 10^{ -2 } \right) =1.1-\cfrac { 0.059 }{ 2 } \times (2)=1.1-0.059=1.041V$

Question 4

The conductivity of 0.20 M solution of $KCl$ at 298 K is $0.0248 S cm^{-1}$. Calculate its molar conductivity.

Accepted Answer

$k$ (Conductivity) $=0.0248 \ Scm^{-1}$$C$ (molarity) $=0.20 \ M$$T$ (temperature) $=298 \ K$$ \Lambda _m =?$ (molar conductivity) The formula is $\Lambda_m = \cfrac {1000 \times k}{C}$$= \cfrac {1000 \times 0.0248 \ Scm^{-1}}{0.20 \ mol\ cm^{-3}}$$=124 \ S \ cm^2 \ mol^{-1}$The molar conductivity is $124 \ S \ cm^2 \ mol^{-1}$

Question 5

Depict the galvanic cell in which the reaction $Zn(s) + 2Ag^{+} (aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$ 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.

Accepted Answer

The galvanic cell in which the given reaction takes place is depicted as: $\displaystyle Zn(s)| Zn^{2+} (aq) || Ag^+ (aq) |Ag(s)$(i) The negatively charged electrode is the zinc electrode. It acts as an anode.(ii) In the external circuit, the current will flow from silver to zinc. (iii) Oxidation at anode: $\displaystyle Zn(s) \rightarrow Zn^{2+}(aq) +2e^-$Reduction at cathode: $\displaystyle Ag^+(aq) + e^- \to Ag(s) $

Question 6

Predict the products of electrolysis in each of the following:(i) An aqueous solution of $AgNO_{3}$ with silver electrodes.(ii) An aqueous solution of $AgNO_{3}$ with platinum electrodes.(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes.(iv) An aqueous solution of $CuCl_{2}$  with platinum electrodes

Accepted Answer

(i) An aqueous solution of $AgNO_{3}$ with silver electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Silver anode will dissolve to form silver ions in the solution.$\displaystyle Ag \rightarrow Ag^+ + e^-$(ii) An aqueous solution of $AgNO_{3}$ with platinum electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-$(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes.At cathode: $\displaystyle  2H^+ + 2e^- \rightarrow H_2(g)$At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-$(iv) An aqueous solution of $CuCl_{2}$  with platinum electrodesAt cathode: Cupric ions will be reduced in preference to protons$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu$At anode: Chloride ions will be oxidized in preference to hydroxide ions$\displaystyle 2Cl^- \rightarrow Cl_2 + 2e^-$

Question 7

Resistance and conductivity of a cell containing 0.001 M KCl solution at 298 K are $1500 \Omega$ and $1.46 \times 10^{-4} S.cm^{-1}$ respectively. What is the cell constant ?

Accepted Answer

$R=1500 \ \Omega$ (Resistance)$k= 1.46 \times 10^{-4} \ Scm^{-1}$ (Conductivity) $T=298 \ K$We will use the formula, $\boxed {k=Gx}$where $k$ is conductivity$G$ is $\cfrac 1R$$x$ is cell constant$1.46 \times 10^{-4} = \cfrac 1{1500} \times x$$1.46 \times 10^{-4} \times 1500 =x$$\boxed {0.219 \ cm^{-1}=x}$The cell constant is $0.219 \ cm^{-1}$

Question 8

In electrolysis of water: (a) Name the gas collected at cathode and anode.(b) Why is volume of one gas collected at one electrode is double of anode?(c) Why are few drops of dilute $H_2SO_4$ added to water?

Accepted Answer

a) At cathode, $H_2$ is evolved and at anode $O_2$ is evolved.b) At cathode, $H^+$ ion takes two electron to convert itself into $H_2$ gas. $2$ moles of $H^+$ gives $1$ mole of $H_2$.$2H^+(aq)+ 2e^- \rightarrow H_2(g)$At anode, $OH^-$ ion releases two electrons to convert into water and $O_2$. $2$ mole $OH^-$ gives $0.5$ mole $O_2$, i.e why volume of one gas collected at one electron is double of another.c) Acid is added to make the water conduct electricity as the distilled water is a non-conductor of electricity.

Question 9

How many electrons are there in one coulomb of electricity?

Accepted Answer

Change on $1e^-=1.602\times 10^{-19}$ $C$If total charge $=1$ $C$$\therefore$ Total no. of $e^-=\cfrac{1}{1.602\times 10^{-19}}=6.24\times 10^{18}e^-$Hence, the correct option is $(C)$.

Question 10

A metal carbonate X on reacting with an acid gives a gas which when passed through a solution gives the carbonate black. On the other hand, A gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives compound Z, used for disinfecting water. Identify X, Y, G and Z.

Accepted Answer

According to the problem:Here,$X$ is calcium carbonate $\left( {CaC{O_3}} \right)$.$Y$ is slaked lime.$Z$ is bleaching powder $\left( {CaOC{l_2}} \right)$$G$ is chlorine gas. Hence the reactions are involved,