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: How would you determine the standard electrode potential of the \$Mg^{2+}\$ \$|Mg\$?

To determine the electrode potential of \$mg^{+2}/mg^{1}\$ we use hydrogen electrode (anode)The standard hydrogen electrode potential is zero.\$\therefore E^{o}cell = E^{o} right+ E^{o} left\$\$E^{o} cell= E^{o} mg^{+2}/mg^{-1} E^{o}n_{2}/n_{1}\$ \$E^{o} mg^{+2}/mg= E^{o} cell\$

Q: How much electricity in terms of Faraday is required to produce:(i) \$20\ g\$ of \$Ca\$ from molten \$CaCl_{2}\$(ii) \$40\ g\$ of \$Al\$ from molten \$Al_{2}O_{3}\$[Given : Molar mass of calcium & Aluminium are 40 g \$mol^{-1}\$ & 27 g \$mol^{-1}\$ respectively]

\$(1)\$ \$\displaystyle Ca^{2+} + 2e^- \rightarrow Ca\$\$1\$ mole (\$40\$ g) of \$Ca\$ will require \$2\$ F of electricity.\$20\$ g (\$0.5\$ mole) of \$Ca\$ will require \$1\$ F of electricity.(2) \$\displaystyle Al^{3+} + 3e^- \rightarrow Al \$\$1\$ mole (\$27\$ g) of \$Al\$ will require \$3\$ moles of electrons.\$40\$ g of \$Al\$ will require \$\displaystyle \dfrac {3 \times 40F}{27} = 4.4\$ F

Q: A solution of \$Ni(NO_{3})_{2}\$ is electrolysed between platinum electrodes using a current of \$5\$ amperes for \$20\$ minutes. What mass of \$Ni\$ is deposited at the cathode?

Given\$I = 5A\$\$Time = 20 \times 60 = 1200\,\,s\$\$\therefore Charg e = current \times time\$\$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 \times 1200\$\$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6000\,\,C\$According to the reaction.\$N{i^{2 + }}\left( {aq.} \right) + 2e \to Ni\,\left( s \right)\$\$Nickel\,\,deposite\,by\, (2 \times 96487)\,C = 58.7\,\,g\$\$\therefore Nickel\,deposite\,by\ 6000\,C = \dfrac{{58.7 \times 6000}}{{2 \times 96487}}\$\$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad= 1.825\,\,g\$\$\therefore \$ Hence \$1.825\,g\$ of nickel will be deposited at the cathode.

Q: During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:

Explanation:We have to use the formula \$W = \frac E {96500} \times I \times t\$.Consider formation of chlorine molecule as shown below:\$2Cl^- \rightarrow Cl_2 + 2e^-\$using formula,\$0.1 \times 71\$ (Molecular mass of \${Cl}_2\$) \$=\frac {35.5}{96500} \times 3 \times t\$.\$\therefore\$ \$t= 6433.33\ seconds \$\$\therefore\$ \$t\approx\$ \$110\ minutes\$Final Answer: Hence option \$D\$ is correct.

Q: Three electrolytic cells \$A, B, C\$ containing solutions of \$ZnSO_{4}, AgNO_{3}\$ and \$CuSO_{3}\$, 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?

Cell \$B\$: \$\displaystyle Ag^+ + e^- \rightleftharpoons Ag\$ at cathode.\$1\$ mole (\$108\$ g) of Ag is deposited by \$96500\$ \$C\$.\$1.45\$ g of Ag will be deposited by \$\displaystyle \dfrac {96500 \times 1.45}{108} = 1295.6 \: C\$.Now,\$\displaystyle Q = It \$\$\displaystyle 1295.6 = 1.5 \times t \$\$\displaystyle t = 864 \: s\$Cell \$A\$: \$\displaystyle Zn^{2+} + 2e^- \rightarrow Zn\$\$2\$ moles of electrons (\$\displaystyle 2 \times 96500\$ C of current) produces \$1\$ mole (\$63.5\$ g) of zinc.\$1295.6\$ \$C\$ of electricity will deposit \$\displaystyle \dfrac {65.3}{2 \times 96500} \times 1295.6 = 0.438\$ g of zincCell \$C\$: \$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu\$\$2\$ moles of electrons (\$\displaystyle 2 \times 96500\$ \$C\$) of current will produce \$1\$ mole (\$63.5\$ g) of \$Cu\$. \$1295.6\$ \$C\$ of current will deposit \$\displaystyle \dfrac {63.5 \times 1295.6}{2 \times 96500} = 0.426 \: g\$ of copper

Q: How much charge is required for the following reductions?(i) \$1\ mol\$ of \$Al^{3+}\$ to \$Al\$(ii) \$1\ mol\$ of \$Cu^{2+}\$ to \$Cu\$(iii) \$1\ mol\$ of \${MnO_4}^{-}\$ to \$Mn^{2+}\$

\$Al^{3+} + 3e^-\rightarrow Al\$The reduction will require 3 Faraday of electricity or \$\displaystyle 3 \times 96500 = 2.895 \times 10^5 \: C\$.\$Cu^{2+} + 2e^-\rightarrow Cu\$The reduction will require 2 Faraday of electricity or \$\displaystyle 2 \times 96500 = 1.93 \times 10^5 \: C\$.\${MnO_4}^{-} + 5e^- \rightarrow Mn^{2+} \$The reduction will require 5 Faraday of electricity or \$\displaystyle 5 \times 96500 = 4.825 \times 10^5 \: C\$.Note: 1 mole of electron has a charge of 1 faraday.

Q: How much electricity is required in coulomb for the oxidation of:(i) \$1\ mol\$ of \$H_{2}O\$ to \$O_{2}\$?(ii) \$1\ mol\$ of \$FeO\$ to \$Fe_{2}O_{3}\$?

(i) \$\displaystyle H_2O \rightarrow 2H^+ + \dfrac{1}{2}O_2 + 2e^-\$Oxidation of one mole of water will require \$\displaystyle 2 \times 96500 = 1.93 \times 10^5 \: C\$.(ii) \$\displaystyle Fe^{2+} \rightarrow Fe^{3+} \$Oxidation of one mole of \$\displaystyle Fe^{2+} \$ will require \$\displaystyle 1 \times 96500 = 96500 \: C\$\$\displaystyle \$.

Q: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Q=ItWhere Q= no. of electronI= currentt= time taken\$Q=0.5 A\times 72005\$\$=3600 C\$As \$96487 C = 6.023\times 10^{23}\$ no. of electron\$\Rightarrow 3600 C = \frac{6.023\times 10^{23}\times 3600}{96487}=2.25\times 10^{22}\$ no. of electron

Question 1

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 2

How would you determine the standard electrode potential of the $Mg^{2+}$ $|Mg$?

Accepted Answer

To determine the electrode potential of $mg^{+2}/mg^{1}$ we use hydrogen electrode (anode)The standard hydrogen electrode potential is zero.$\therefore E^{o}cell = E^{o} right+ E^{o} left$$E^{o} cell= E^{o} mg^{+2}/mg^{-1} E^{o}n_{2}/n_{1}$ $E^{o} mg^{+2}/mg= E^{o} cell$

Question 3

How much electricity in terms of Faraday is required to produce:

(i) $20 g$ of $C a$ from molten $C a C l_{2}$
(ii) $40 g$ of $A l$ from molten $A l_{2} O_{3}$

[Given : Molar mass of calcium & Aluminium are 40 g $m o l^{- 1}$ & 27 g $m o l^{- 1}$ respectively]

Accepted Answer

$(1)$ $\displaystyle Ca^{2+} + 2e^- \rightarrow Ca$$1$ mole ($40$ g) of $Ca$ will require $2$ F of electricity.$20$ g ($0.5$ mole) of $Ca$ will require $1$ F of electricity.(2) $\displaystyle Al^{3+} + 3e^- \rightarrow Al $$1$ mole ($27$ g) of $Al$ will require $3$ moles of electrons.$40$ g of $Al$ will require $\displaystyle \dfrac {3 \times 40F}{27} = 4.4$ F

Question 4

A solution of $Ni(NO_{3})_{2}$ is electrolysed between platinum electrodes using a current of $5$ amperes for $20$ minutes. What mass of $Ni$ is deposited at the cathode?

Accepted Answer

Given$I = 5A$$Time = 20 \times 60 = 1200\,\,s$$\therefore Charg e = current \times time$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 \times 1200$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6000\,\,C$According to the reaction.$N{i^{2 + }}\left( {aq.} \right) + 2e \to Ni\,\left( s \right)$$Nickel\,\,deposite\,by\, (2 \times 96487)\,C = 58.7\,\,g$$\therefore Nickel\,deposite\,by\ 6000\,C = \dfrac{{58.7 \times 6000}}{{2 \times 96487}}$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad= 1.825\,\,g$$\therefore $ Hence $1.825\,g$ of nickel will be deposited at the cathode.

Question 5

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:

Accepted Answer

Explanation:We have to use the formula $W = \frac E {96500} \times I \times t$.Consider formation of chlorine molecule as shown below:$2Cl^- \rightarrow Cl_2 + 2e^-$using formula,$0.1 \times 71$ (Molecular mass of ${Cl}_2$) $=\frac {35.5}{96500} \times 3 \times t$.$\therefore$ $t= 6433.33\ seconds $$\therefore$ $t\approx$ $110\ minutes$Final Answer: Hence option $D$ is correct.

Question 6

Three electrolytic cells $A, B, C$ containing solutions of $ZnSO_{4}, AgNO_{3}$ and $CuSO_{3}$, 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?

Accepted Answer

Cell $B$: $\displaystyle Ag^+ + e^- \rightleftharpoons Ag$ at cathode.$1$ mole ($108$ g) of Ag is deposited by $96500$ $C$.$1.45$ g of Ag will be deposited by $\displaystyle \dfrac {96500 \times 1.45}{108} = 1295.6 \: C$.Now,$\displaystyle Q = It $$\displaystyle 1295.6 = 1.5 \times t $$\displaystyle t = 864 \: s$Cell $A$: $\displaystyle Zn^{2+} + 2e^- \rightarrow Zn$$2$ moles of electrons ($\displaystyle 2 \times 96500$ C of current) produces $1$ mole ($63.5$ g) of zinc.$1295.6$ $C$ of electricity will deposit $\displaystyle \dfrac {65.3}{2 \times 96500} \times 1295.6 = 0.438$ g of zincCell $C$: $\displaystyle Cu^{2+} + 2e^- \rightarrow Cu$$2$ moles of electrons ($\displaystyle 2 \times 96500$ $C$) of current will produce $1$ mole ($63.5$ g) of $Cu$. $1295.6$ $C$ of current will deposit $\displaystyle \dfrac {63.5 \times 1295.6}{2 \times 96500} = 0.426 \: g$ of copper

Question 7

How much charge is required for the following reductions?(i) $1\ mol$ of $Al^{3+}$ to $Al$(ii) $1\ mol$ of $Cu^{2+}$ to $Cu$(iii) $1\ mol$ of ${MnO_4}^{-}$ to $Mn^{2+}$

Accepted Answer

$Al^{3+} + 3e^-\rightarrow Al$The reduction will require 3 Faraday of electricity or $\displaystyle 3 \times 96500 = 2.895 \times 10^5 \: C$.$Cu^{2+} + 2e^-\rightarrow Cu$The reduction will require 2 Faraday of electricity or $\displaystyle 2 \times 96500 = 1.93 \times 10^5 \: C$.${MnO_4}^{-} + 5e^- \rightarrow Mn^{2+} $The reduction will require 5 Faraday of electricity or $\displaystyle 5 \times 96500 = 4.825 \times 10^5 \: C$.Note: 1 mole of electron has a charge of 1 faraday.

Question 8

The weight of silver (atomic weight $= 1 08$) displaced by a quantity of electricity which displaces 5600 ml of $O_2$ at STP will be:

Accepted Answer

At STP, 1 mole of oxygen occupies 22400 mL.Hence, the number of moles of oxygen corresponding to 5600 ml is $n_{O_2}=\dfrac {5600}{22400}=\dfrac {1}{4}=\dfrac {W_{O_2}}{M_{O_2}}$.1 mole of oxygen produces 4 moles of lectrons and 1 mole of silver requires 1 mole of electrons.$\dfrac {W_{Ag}}{M_{Ag}}\times 1=\dfrac {W_{O_2}}{M_{O_2}}\times 4$ $\dfrac {W_{Ag}}{M_{Ag}}\times 1=\dfrac {1}{4}\times 4$ $\dfrac {W_{Ag}}{108}=\dfrac {1}{4}\times 4$$W_{Ag}=108 g$    Thus, the weight of silver displaced by a quantity of electricity which displaces 5600 ml of $O_2$ at STP will be 108.0 g.

Question 9

How much electricity is required in coulomb for the oxidation of:(i) $1\ mol$ of $H_{2}O$ to $O_{2}$?(ii) $1\ mol$ of $FeO$ to $Fe_{2}O_{3}$?

Accepted Answer

(i) $\displaystyle H_2O \rightarrow 2H^+ + \dfrac{1}{2}O_2 + 2e^-$Oxidation of one mole of water will require $\displaystyle 2 \times 96500 = 1.93 \times 10^5 \: C$.(ii) $\displaystyle Fe^{2+} \rightarrow Fe^{3+} $Oxidation of one mole of $\displaystyle Fe^{2+} $ will require $\displaystyle 1 \times 96500 = 96500 \: C$$\displaystyle $.

Question 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?

Accepted Answer

Q=ItWhere Q= no. of electronI= currentt= time taken$Q=0.5 A\times 72005$$=3600 C$As $96487 C = 6.023\times 10^{23}$ no. of electron$\Rightarrow 3600 C = \frac{6.023\times 10^{23}\times 3600}{96487}=2.25\times 10^{22}$ no. of electron

Revise with Concepts

Electrolytic Cells and Electrolysis

Faraday's Law

Products of Electrolysis

Quick Summary With Stories

Introduction to Electrolysis

Faraday Second Law of Electrolysis

Calculate the emf of the cell in which of the following reaction takes place:

$N i (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 A g (s)$

[Given that $E_{c e l l}^{\circ} = 1.05 V$ ]

How would you determine the standard electrode potential of the $M g^{2 +}$ $∣ M g$ ?

How much electricity in terms of Faraday is required to produce:

(i) $20 g$ of $C a$ from molten $C a C l_{2}$
(ii) $40 g$ of $A l$ from molten $A l_{2} O_{3}$

[Given : Molar mass of calcium & Aluminium are 40 g $m o l^{- 1}$ & 27 g $m o l^{- 1}$ respectively]

A solution of $N i (N O_{3})_{2}$ is electrolysed between platinum electrodes using a current of $5$ amperes for $20$ minutes. What mass of $N i$ is deposited at the cathode?

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:

How much charge is required for the following reductions?

(i) $1 m o l$ of $A l^{3 +}$ to $A l$
(ii) $1 m o l$ of $C u^{2 +}$ to $C u$
(iii) $1 m o l$ of $M n O_{4}^{-}$ to $M n^{2 +}$

The weight of silver (atomic weight $= 108$ ) displaced by a quantity of electricity which displaces 5600 ml of $O_{2}$ at STP will be:

How much electricity is required in coulomb for the oxidation of:
(i) $1 m o l$ of $H_{2} O$ to $O_{2}$ ?
(ii) $1 m o l$ of $F e O$ to $F e_{2} O_{3}$ ?

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Revise with Concepts

Electrolytic Cells and Electrolysis

Faraday's Law

Products of Electrolysis

Quick Summary With Stories

Introduction to Electrolysis

Faraday Second Law of Electrolysis

Calculate the emf of the cell in which of the following reaction takes place: Ni(s)+2Ag+(0.002M)→Ni2+(0.160M)+2Ag(s) [Given that Ecell∘​=1.05V]

How would you determine the standard electrode potential of the Mg2+ ∣Mg?

How much electricity in terms of Faraday is required to produce: (i) 20 g of Ca from molten CaCl2​ (ii) 40 g of Al from molten Al2​O3​ [Given : Molar mass of calcium & Aluminium are 40 g mol−1 & 27 g mol−1 respectively]

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?

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:

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 MnO4​− to Mn2+

The weight of silver (atomic weight =108) displaced by a quantity of electricity which displaces 5600 ml of O2​ at STP will be:

How much electricity is required in coulomb for the oxidation of: (i) 1 mol of H2​O to O2​? (ii) 1 mol of FeO to Fe2​O3​?

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Calculate the emf of the cell in which of the following reaction takes place:

$N i (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 A g (s)$

[Given that $E_{c e l l}^{\circ} = 1.05 V$ ]

How would you determine the standard electrode potential of the $M g^{2 +}$ $∣ M g$ ?

How much electricity in terms of Faraday is required to produce:

(i) $20 g$ of $C a$ from molten $C a C l_{2}$
(ii) $40 g$ of $A l$ from molten $A l_{2} O_{3}$

[Given : Molar mass of calcium & Aluminium are 40 g $m o l^{- 1}$ & 27 g $m o l^{- 1}$ respectively]

A solution of $N i (N O_{3})_{2}$ is electrolysed between platinum electrodes using a current of $5$ amperes for $20$ minutes. What mass of $N i$ is deposited at the cathode?

How much charge is required for the following reductions?

(i) $1 m o l$ of $A l^{3 +}$ to $A l$
(ii) $1 m o l$ of $C u^{2 +}$ to $C u$
(iii) $1 m o l$ of $M n O_{4}^{-}$ to $M n^{2 +}$

The weight of silver (atomic weight $= 108$ ) displaced by a quantity of electricity which displaces 5600 ml of $O_{2}$ at STP will be:

How much electricity is required in coulomb for the oxidation of:
(i) $1 m o l$ of $H_{2} O$ to $O_{2}$ ?
(ii) $1 m o l$ of $F e O$ to $F e_{2} O_{3}$ ?