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

Products of Electrolysis | Definition, Examples, Diagrams

definition

Factors affecting Electrolysis

Factors Affecting Products of Electrolysis are as follows:
1. Products of electrolysis depend on the material being electrolyzed. In other words, the nature of electrolyte governs the process of electrolysis. The process is fast for a strong electrolyte whereas for weak electrolyte an extra potential better known as overpotential is required. Products of electrolysis depend on upon the value of this overpotential too.
2. Products of electrolysis depend on the nature of electrodes too. That is, in the case of the inert electrode (say gold, platinum), it doesnt participate in the reaction whereas if the electrode used is reactive in nature it takes part in the reaction.
3. Various oxidising and reducing species present in the electrolytic cell do affect the products of electrolysis.
4. The products of electrolysis depend on standard electrode potentials of the different oxidizing and reducing species present in the electrolytic cell.
5. In case of multiple reactions, product of electrolysis depends on the standard electrode potential of various reactions taking place. Out of the multiple reduction reactions taking place, the reduction reaction which has highest value of standard electrode potential takes place at cathode. Similarly, out of the multiple oxidation reactions, the oxidation reaction which has the lowest value of standard electrode potential takes place at anode.

example

Electrolysis of Molten NaCl

If sodium chloride is melted (above 801 C), two electrodes are inserted into the melt, and an electric current is passed through the molten salt, then chemical reactions take place at the electrodes. Since the electrolyte is molten Sodium Chloride we only have

N a^{+}

and

C l^{-}

ions in the solution.
Sodium ions migrate to the cathode, where electrons enter the melt and are reduced to sodium metal:

N a^{+} + e^{-} \to N a

Chloride ions migrate the other way, toward the anode. They give up their electrons to the anode and are oxidized to chlorine gas:

C l^{-} \to \frac{1}{2} C l_{2} + e^{-}

The overall reaction is the breakdown of sodium chloride into its elements:

2 N a C l \to 2 N a_{(s)} + C l_{2} (g)

definition

Electrolysis of Aqueous CuBr

When we have an aqueous solution of Copper Bromide we also have to take water into the equation. Since water can be both oxidized and reduced, it competes with the dissolved

C u^{2 +}

and

B r^{-}

ions. So the ions present in the solution are:

B r^{-}

C u^{2 +}

H^{+}

O H^{-}

.
The positive ions are attracted to the negative cathode. There is competition between the copper ions and the hydrogen ions. As the hydrogen ion has lower reduction potential than the copper ion (

E_{R}^{o}

=+0.521V), the copper ions are preferentially reduced and copper metal is deposited at the electrode (a pink layer is observed)
The reaction at Cathode is:

C u_{2 +} + 2 e^{-} \to C u (s)

The reaction at anode is:

2 B r^{-} \to B r_{2 (g)} + 2 e^{-}

Reduction of

C u^{2 +}

(

E^{o}

= 0.521 V) is energetically more favorable than the reduction of

H_{2} O

(

E^{o}

= 0 V). So in aqueous solution, reduction of Copper will take place.
At the anode, where oxidation occurs, the standard oxidation potential of water is -1.23 volts, while that for bromide ions is -1.07 volts. Hence, Oxidation of Bromine will take place, which is energetically more favorable.

definition

Electrolysis of Aqueous NaCl

When we have an aqueous solution of sodium chloride we also have to take water into the equation. Since water can be both oxidized and reduced, it competes with the dissolved

N a^{+}

and

C l^{-}

ions. Rather than producing sodium, hydrogen is produced.
The reaction at Cathode is:

H_{2} O_{(l)} + 2 e^{-} \to H_{2 (g)} + 2 O H^{-}

The reaction at anode is:

C l^{-} \to \frac{1}{2} C l_{2 (g)} + 1 e^{-}

Reduction of

N a^{+}

(

E^{o}

= -2.7 V) is energetically more difficult than the reduction of water (

E^{o}

= 0.0 V). So in aqueous solution, reduction of water will take place.

Products of Electrolysis

definition

Factors affecting Electrolysis

example

Electrolysis of Molten NaCl

definition

Electrolysis of Aqueous CuBr

definition

Electrolysis of Aqueous NaCl

REVISE WITH CONCEPTS

Electrolytic Cells and Electrolysis

Faraday's Law

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?

definition

Factors affecting Electrolysis

example

Electrolysis of Molten NaCl

definition

Electrolysis of Aqueous CuBr

definition

Electrolysis of Aqueous NaCl

REVISE WITH CONCEPTS

Electrolytic Cells and Electrolysis

Faraday's Law

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