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}\$

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\$

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

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.

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.

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

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.

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:

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.

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 \$.

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

Faraday's Law | Formulas, Definition, Law

formula

Electrical Units of Charge

Following are the units of charge.

Coulomb $(C)$ : Coulomb is the smaller unit of charge or electricity . The amount of charge that passes a given point when 1A current flows for 1 s.
Faraday $(F)$ : The amount of electric charge on one mole of electron is called as Faraday.

law

Faraday's Law of Electrolysis

Faraday's First Law of Electrolysis
The amount of substance that undergoes a chemical reaction at an electrode during electrolysis is proportional to the quantity of electricity passed through an electrolyte.

W \propto Q

\Rightarrow W \propto I t

\Rightarrow W = k I t

W

is the weight of the substance

Q

is the amount of charge passed

I

is current

t

is time for which current flows

k

is proportional constant.

Faraday's Second Law of Electrolysis
The mass of a substance deposited or liberated at any electrode on passing a certain amount of charge is directly proportional to its chemical equivalent weight.

\frac{W _{1}}{W _{2}} = \frac{E _{1}}{E _{2}}

Where,

W_{1}

and

W_{2}

are weight deposited of two elements 1 and 2 respectively

E_{1}

and

E_{2}

are the equivalent weights of two elements 1 and 2 respectively.

example

Numerical on Faraday's First Law

Question: When

1000 C

of charge is passed through

C u S O_{4}

solution,

x

g of copper is deposited. How much charge should be passed through the solution to deposit

5 x

g of copper?
Solution: According to Faraday's First Law,

W = k Q

First Case:

x = k \times 1000 C

Second Case:

5 x = k \times y

y

is a charge to deposit

5 x

g of copper.
Dividing first case by second one
We get,

y = 5000 C

law

Faraday's Second Law of Electrolysis

Faraday's Second Law of Electrolysis states that the mass of a substance deposited or liberated at any electrode on passing a certain amount of charge is directly proportional to its chemical equivalent weight.

\frac{W _{1}}{W _{2}} = \frac{E _{1}}{E _{2}}

Where,

W_{1}

and

W_{2}

are weight deposited of two elements 1 and 2 respectively

E_{1}

and

E_{2}

are the equivalent weights of two elements 1 and 2 respectively.

result

Combination of Two Faraday's Law

W = \frac{E}{F} I t

By the passage of 1 mole of electrons

(1 F = 96500 C)

, the equivalent weight of a substance is deposited or liberated.
This is a combined Faraday's Law of Electrolysis.

example

Numerical on Faraday's Second Law

Question: A certain quantity of electricity is passed through an aqueous solution of

A g N O_{3}

and cupric salt solution connected in series. The amount of silver deposited is

1.08

g. The amount of copper deposited is:
(Atomic wt. of

C u = 63.54, A g = 108

g/mole)
Solution: Equivalent weight of silver

= 108

g/eq
Equivalent weight of copper

= 31.77

g/eq
Mass of

A g

deposited

= 1.08

g
Substitute values in the following equation,

\frac{M a s s o f C u d e p o s i t e d}{M a s s o f A g d e p o s i t e d} = \frac{E q . W t o f C u}{E q . W t o f A g}

\frac{M a s s o f C u d e p o s i t e d}{1 . 0 8} = \frac{3 1 . 7 7}{1 0 8}

M a s s o f C u d e p o s i t e d = \frac{3 1 . 7 7}{1 0 8} \times 1.08 = 0.3177

definition

Define Electrolysis

Electrolysis refers to the decomposition of a substance by an electric current.

Example: When the current is passed through the molten sodium chloride, sodium and chlorine are deposited at different electrodes. Thus, sodium chloride is decomposed into sodium and chlorine

definition

Define and explain electrolysis

Electrolysis is the process by which electric current is passed through a substance to effect a chemical change. The chemical change is one in which the substance loses or gains an electron (oxidation or reduction). The process is carried out in an electrolytic cell, an apparatus consisting of positive and negative electrodes held apart and dipped into a solution containing positively and negatively charged ions. The substance to be transformed may form the electrode, may constitute the solution, or may be dissolved in the solution. Electric current (i.e., electrons) enters through the negatively charged electrode (cathode); positively charged components of the solution travel to this electrode, combine with the electrons, and are transformed to neutral elements or molecules. The negatively charged components of the solution travel to the other electrode (anode), give up their electrons, and are transformed into neutral elements or molecules. If the substance to be transformed is the electrode, the reaction is generally one in which the electrode dissolves by giving up electrons.

Faraday's Law

formula

Electrical Units of Charge

law

Faraday's Law of Electrolysis

example

Numerical on Faraday's First Law

law

Faraday's Second Law of Electrolysis

result

Combination of Two Faraday's Law

example

Numerical on Faraday's Second Law

definition

Define Electrolysis

definition

Define and explain electrolysis

REVISE WITH CONCEPTS

Electrolytic Cells and Electrolysis

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?

formula

Electrical Units of Charge

law

Faraday's Law of Electrolysis

example

Numerical on Faraday's First Law

law

Faraday's Second Law of Electrolysis

result

Combination of Two Faraday's Law

example

Numerical on Faraday's Second Law

definition

Define Electrolysis

definition

Define and explain electrolysis

REVISE WITH CONCEPTS

Electrolytic Cells and Electrolysis

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 $F e_{2} O_{3}$ ?