Chemistry NCERT Exemplar Solutions Class 12th Chapter Three Electrochemistry

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$\therefore n_x=1$

$n_y=1$

$n_z=\frac{0.05}{3}=0.0167$  here z is limiting reagent.

$?\frac{0.05}{3}$  mole z gives 1 mole xyz₃

mass of xyz₃ = n × molecular mass

=  $\frac{0.05}{3}\times \left(10+20+3\times 30\right)a.m.u.$  

= 0.5 × 4 = 2g

B.C.C structure

              a = 300 pm = 300 × 10^-12 m

              d = 6g/cm³

              z = 2

              $d=\frac{Z\times M}{a^3}$  

              $6=\frac{2\times A}{{\left(300\times 10^{-10}\right)}^3}=\frac{2\times A}{27\times 10^{-24}}$  

              $\therefore AtomsofM$  = 3.69 × 6.022 × 10²³

                                           = 22.22 × 10²³

              the nearest integer = 22

$B_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1={\pi }_{2py}^1\to$ Paramagnetic

$\begin{array}{l}L{i}_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2\to Diamagnetic\\ C_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2\to Diamagnetic\\ C_2^{-}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^1\to Paramagnetic\end{array}$

$O_2^{-2}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*2}\equiv {\pi }_{2py}^{*2}$ Diamagnetic

$O_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^0\to$ Paramagnetic

$H{e}_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*1}\to$ Paramagnetic

$\therefore$ Paramagnetic molecules are $=B_2,C_2^{-},O_2^{+},H{e}_2^{+}$

$m=\frac{w\times 1000}{molecularwt\times w_{solvent}}=\frac{10.2\times 1000}{176\times 150}$

$=\frac{1020\cancel{0}}{176\times 15\cancel{0}}=0.386$

$\Delta T_f=K_f\times m$

3.9 × 0.386

$\therefore x\times 10^{-1}=15.05\times 10^{-1}$

Ans = 15

Ka for C₃H₇COOH = 2 × 10^-5

              $pKa=-\mathcal{l}og\left(2\times 10^{-5}\right)=5-\mathcal{l}og2$  

              =5 – 0.3 = 4.7

              pH of 0.2 (M) solution =

              $pH=\frac{pKa-\mathcal{l}ogC}{2}$  

              $=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

              $pH=27\times 10^{-1}$     

          $\therefore$    Ans 27

t_1/2 = 0.301 min

              t = 2 min

              $K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{2.303\times 0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{Co}{Ct}=10^2=100$  

              Ans. 100

$C\mathcal{l}{F}_3$ ® T-shaped (sp³d)

 IF₇ ® Pentagonal bipyramidal (sp³d³)

BrF₅ ® Square pyramidal (sp³d²)

BrF₃ ® T-Shaped (sp³d)

I₂Cl₆ ® Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF ® (sp³)

ClF₅ ® Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

$Mn{O}_4^{-2}\to A+B$  

              Oxidation state of Mn in B < A

              $Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

              B is MnO₂

              $\therefore$  Oxidation state of Mn = +4

              ${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

              $\therefore$  unpaired electron = 3

              Spin only magnetic moment $\left(\mu \right)$  

              $=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

              $=3.87\approx 4$  

$\left[Cu{\left(en\right)}_2{\left(SCN\right)}_2\right]$

More stable isomers = 3 (trans isomers)

% of C in organic compound

              $=\frac{12}{14}\times \frac{WC{O}_2}{W.O.C.}\times 100$  

              $=\frac{952.56}{21.648}=44\mathrm{\%}$  

              $\mathrm{\%}ofH=\frac{2}{18}\times \frac{W{H}_{2}O}{W.O.C.}\times 100$  

              = $\frac{20}{18}\times \frac{0.4428}{0.492}\times 100$  

              = $\frac{88.56}{8.856}=10\mathrm{\%}$  

              = 100 – 54 = 46%

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Ans: Alternating current is used to stop electrolysis so that the concentration of the ions in the solution remains constant.

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Ans:  Galvanic cell consists of two electrodes in which anode and cathode are present as electrodes. The anode is present on the left side at which oxidation occurs. The cathode is present on the right side at which reduction occurs and in the middle, there is a salt bridge which is depicted by parallel lines. Therefore, the galvanic cell is Cu|Cu²⁺ |Ag⁺ |Ag.

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Ans: The potential difference between the metal and its solution is termed electrode potential.

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Ans: Conductivity due to the total number of ions present in the solution is termed as specific conductivity. An increase in the number of ions per unit volume increases. In addition to water, the number of ions per unit volume decreases therefore conductivity decreases.

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Ans: Standard hydrogen electrode (SHE) is used as a reference electrode and its electrode potential is assumed to be zero. The electrode potential of other electrodes is measured concerning standard hydrogen electrodes.

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Ans: The given cell reaction consists of two half cell reactions represented below as follows;

At anode: Cu→Cu²⁺+ 2e⁻

At cathode: Cl² + 2e⁻ →2Cl⁻

Here oxidation of copper and reduction of chlorine is taking place

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Ans: According to Faraday’s second law of electrolysis, the amount of different substances liberated keeping the electricity flow same through the electrolytic solution is directly proportional to the equivalent weight.

$\frac{{\mathrm{\omega }}_1}{{\mathrm{\omega }}_2}$ = $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}$

E₁and E₂ have different values. Therefore, the mass of copper and silver deposited will be different

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Ans: In the electrolysis process of sodium chloride, oxidation of water at anode requires more potential. So Cl− is oxidized at anode instead of water.

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Ans: The cell represented above shows an electrochemical cell in which two different electrodes are present and the cell at the bottom represents the electrolytic cell.

In this zinc is losing electrons moving towards electrode A and copper is accepting an electron from electrode B. Therefore, the polarity of electrode A is positive and electrode B is negative.

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Ans: When the opposing potential becomes equal to the electrical potential there is no current flowing in the cell and the cell reaction stops and there is no chemical reaction in the cell

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Ans: Electrolysis of brine solution is as follows

NaCl (aq)→Na⁺ (aq)+Cl⁻ (aq)

At the cathode, the reaction goes this way,

H₂O (l) + e⁻→12H₂ (g)+OH⁻ (aq)

At the anode, the reaction goes this way,

Cl⁻ (aq)→12Cl₂ (g) + e⁻

The pH of the solution will increase as sodium hydroxide is being formed.

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Ans: The number of ions present in the cell decides the life-time of any cell. In mercury cells, ions are not involved so the mercury cell has a constant cell potential throughout its useful life.

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Ans: Stronger electrolytes undergo cent percent dissociation. Electrolyte B will be the stronger electrolyte since the concentration of ions remains the same upon dilution.

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Ans: pH of the substance depends on the concentration of hydronium ions present. Here the concentration of hydronium ions remains constant so pH remains constant.

The electrolysis reaction is as follows;

At anode: 2H₂O (l)→O₂ (g)+ 4H⁺ + 4e−

At cathode:4H⁺ + 4e⁻ →2H₂

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Ans: The reactions taking place in the Daniel cell is as follows;

Zn⁺ + Cu²⁺ → Zn²⁺ + Cu

The number of electrons involved in two. The Nernst equation is as follows;

E_{cell =} E $°$ _cell - $\frac{-0.059}{2}$  log $\frac{\left[Z{n^2}^{+}\right]}{\left[C{u^2}^{+}\right]}$

From the above equation, we conclude that increasing the concentration of Zn^2+, E_cell will decrease.

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Ans: Primary batteries contain a limited amount of reactants and once consumed are discharged. Secondary batteries take a long time to recharge. Fuel cells run continuously as long as reactants are supplied to them and products are consumed

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Ans: The cell reaction of a lead storage battery is as follows;

Pb + PbO₂ + 2H₂SO₄→2H₂SO₄ + 2H₂O

The density of electrolytes changes because water is formed and sulphuric acid is consumed as a product during discharge of the battery

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Ans: Acetic acid is a weak electrolyte and the number of ions on dilution increases due to an increase in the degree of dissociation. In the case of strong electrolytes the number of ions remains the same but interionic attraction decreases.

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Ans: The relation between Gibbs free energy and the emf of the cell is as follows;

ΔG=−nFE_cell

 E $°$ _cell s the cell potential

is the standard emf of the cell

Maximum work obtained from the galvanic cell is nFE $°$ _.

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Ans: (i) The cell reaction of a lead storage battery is as follows;

Pb + PbO₂ + 2H₂SO₄→2H₂SO₄ + 2H₂O

At cathode:PbO₂ (s)+SO₄²⁻ (aq)+2e⁻→2PbSO₄ (s) + 2H₂O (l)

At anode:Pb (s)+SO₄²⁻ (aq)→PbSO₄ (s)+2e⁻

Therefore Pb is the anode and PbO₂ is cathode

(ii)Mercury cell doesn’t contain ions so it gives steady potential.

(iii)Fuel cells have maximum efficiency as they produce energy due to the combustion reaction of fuel.

(iv)Rusting is prevented by galvanization.

Hence, the answer is:

(i)-d ; (ii)-c ; (iii)-a ; (iv)-b

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Ans:   (i)Conductivity k is defined as k = $°$

(ii)Molar conductivity is given by Λ_m = $\times \mathit{T}$

(iii)Degree of dissociation  is given by $\frac{\mathbf{\Lambda }\mathbf{m}}{{\mathbf{\Lambda }}^0\mathbf{m}}$ ? = $\mathit{\alpha }$

(iv)Charge $\frac{\mathit{K}}{\mathit{C}}$  is the product of current and time $\mathit{}\mathit{Q}$  = I $\frac{\mathit{G}}{\mathit{R}}$

Hence, the answer is:

(i)-d; (ii)-c; (iii)-b; (iv)-a

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Ans:  (i)The reaction occurring at Lechlanche cells are 

At anode: Zn→Zn²⁺ + 2e⁻

At cathode: MnO₂ + NH₄⁺ + e⁻→MnO (OH)+NH₃

(ii)Ni−Cd is rechargeable. So it has a longer lifetime.

(iii) Energy in the fuel cell is due to the combustion process. It converts combustion energy into electrical energy 2H₂ + O₂→2H₂O

(iv) Mercury cells don’t involve any solution and are used in hearing aids.

Hence, the answer is:

(i)-d; (ii)-c; (iii)-a, e; (iv)-b

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Ans:   (i) F₂ is a non-metal which is the best oxidizing agent since the standard reduction potential of F₂ is 2.87V.

(ii)Liis a metal and the strongest reducing agent because the standard reduction potential of Li is −3.05V.

(iii)Au³⁺ is a metal ion which is an oxidizing agent as standard reduction potential of Au³⁺ is 1.40V.

(iv) Br⁻is an anion that can be oxidized by Au³⁺ as standard reduction potential of Au³⁺ is 1.40V which is more than that of Br⁻ which is E $\mathit{Q}$ _Br2/Br−0=1.09V.

(v)Auis an unreactive metal

(vi)Li⁺ is a metal ion having the least value of standard reduction potential of 3.05V. Hence it is the weakest oxidizing agent

(vii)F⁻is an anion that is the weakest reducing agent as it has less oxidation potential.

Hence, the answer is:

(i)-c; (ii)-a; (iii)-g; (iv)-e; (v)-d; (vi)-b; (vii)-f

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Correct Option: (i)

Ans: (i) Conductivity is due to the movement of ions. The number of ions present per unit volume decides the conductivity. It decreases on dilution. The number of ions per unit volume decreases on dilution. So, the assertion is correct, the reason is correct and the reason is the correct explanation of the assertion. Therefore, this option is correct

(ii) Assertion is correct, the reason is correct and the reason is the correct explanation of assertion. Therefore, this option is incorrect

(iii) Assertion and true both are correct therefore this option is incorrect

(iv) Assertion is true and the reason is false therefore this option is incorrect

(v) Assertion is true therefore this option is incorrect.

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Ans: Correct Option: (i)

  (i) Molar conductivity of weak electrolytes increases on dilution. In addition to solution the degree of dissociation increases leading to an increase in the number of ions produced. Therefore,   Λm shows a sharp increase.

Therefore, the assertion is correct, the reason is correct and the reason is a correct explanation for assertion. 

(ii) Assertion is correct, the reason is correct and the reason is the correct explanation of assertion. Therefore, this option is incorrect

(iii)Assertion and true both are correct therefore this option is incorrect

(iv) Assertion is true and the reason is false therefore this option is incorrect

(v)Assertion is true therefore this option is incorrect.

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Ans: Correct Option: (i)

(i)Mercury cells give a steady potential. The reason is that the ions are not involved in this reaction. The assertion is incorrect and the reason is correct so this option is incorrect

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Ans: Correct Option: (i)

  (i) Electrolysis of sodium chloride is represented by the following equations

H⁺ (aq)+ e⁻→12H₂ (g)

At the anode, the reaction goes this way,

Cl⁻ (aq)→12Cl₂ (g) + e⁻ E $°$ _cell =1.36V

2H₂O (aq)→O₂ (g) + 4H⁺ (aq)+ 4e− E $°$ _cell =1.23V

The reaction with the lower value E_cell is preferred but the formation of oxygen requires overpotential

Therefore, the assertion is correct, the reason is correct and the reason is the correct explanation for assertion. Therefore, this option is correct

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Ans: Correct Option: (i)

 The resistance of an ionic solution is measured by alternating current. The concentration of DC changes with electrolysis. Therefore, the assertion is correct, the reason is correct and the reason is a correct explanation of the assertion.

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Ans: Correct Option: (i)

(i)At the equilibrium stage E $°$ _cell = 0 and current stops flowing in the cell. Therefore, the assertion is correct, the reason is correct and the reason is the correct explanation for assertion. Therefore, this option is correct

(ii)  Assertion is correct, the reason is correct and the reason is the correct explanation for assertion. Therefore, this option is incorrect

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Ans: Correct Option: (iv)

  (i) Copper sulfate can’t be stored in zinc vessels due to less reactive nature than zinc due to the negative standard reduction value of zinc.

So both assertion and reason are false. Therefore, this option is incorrect.

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Ans: Correct Option: (ii)

  (i) The Nernst equation can be written as follows

E_{cell =} E $°$ _{cell – $\frac{0.059}{1}$   log $\frac{1}{\left[\mathbf{A}{\mathbf{g}}^{+}\right]}$}

E_{cell =} E $°$ _{cell +} 0.059 _logAg+

Here we observe that E_Ag+/Ag increases with increase in the concentration of Ag+ ions. It has a positive value.

Here both assertion and reason are correct .The reason is not a correct explanation of assertion. So this option is incorrect

(ii) The Nernst equation can be written as follows

E_{cell =} E $°$ _{cell – $\frac{0.059}{1}$  log $°$}

E_{cell =} E_{cell +} 0.059 Ag+

Here we observe that E_Ag+/Ag increases with increase in the concentration of Ag+ ions. It has a positive value.

Here both assertion and reason are correct. The correct explanation of assertion is not the reason given. So, this option is correct

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Ans: Correct Option: (ii)

According to the electrochemical series and standard reduction potential of metal, the higher the negative value of standard reduction potential stronger will be the reducing agent

In the given options, the standard reduction potential of chromium has the highest value so it is the strongest reducing agent.

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Ans: Correct Option: (iii)

(i) The properties which don’t depend on the mass of species are termed intensive properties. E_Cell is an intensive property that doesn’t depend on the mass of the species.? _rG is an extensive property as it depends on the mass of the species. Therefore, this option is incorrect

(ii) E_Cell is an intensive property but? _rG is an extensive property so this option is incorrect

(iii) The properties which don’t depend on the mass of species are termed intensive properties E_Cell is an intensive property that doesn’t depend on the mass of the species.? _rG is an extensive property as it depends on the mass of the species. Therefore, this option is correct

(iv)  E_Cell is an intensive property but? _rG is an extensive property so this option is incorrect.

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Ans: Correct Option: (iii)

 The difference between electrode potential potentials of two electrodes when no current is drawn through the cell is termed as cell emf. It is the energy provided by a cell per coulomb of charge passing through it. It is the maximum potential difference between the electrodes of a cell.

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Ans: Correct Option: (iv)

  (i) Inert electrodes like graphite are a source of electrons in a reaction. It doesn't participate in a chemical reaction. Therefore, this option is incorrect

(ii) Inert electrode chemically doesn't participate in the reaction but it provides a surface either for oxidation and reduction reaction. Therefore, this option is incorrect

(iii) Inert electrodes provide surface for conduction of electrons so this option is incorrect.

(iv) Inert electrodes cease transfer of electrons for oxidation or reduction but not for redox reaction. Therefore, this option is correct.

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Ans: Correct Option: (iii)

On the application of external potential, an increased reaction starts till the opposing voltage becomes 1.1V.

However, further increase in the external potential leads the reaction to start in an opposite direction functioning as an electrolytic cell which is a device that uses electrical energy for carrying out electrochemical reactions.

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Ans: Correct Option: (iii)

(i) Conductivity of solution depends upon the size of ions as more is the size of ion, mobility decreases, and conductivity decreases so this option is incorrect.

(ii) Viscosity is the measure of the resistance of a liquid to flow. Greater will be the viscosity of solvent, less will be the flow of electrons and lesser will be the conductivity. Therefore, this option is incorrect.

(iii) Conductivity depends on the solvation of ions in the solution. Greater is the solvation, lesser will be the conductivity of the solution. Therefore, this option is correct.

(iv) On increasing temperature the moving ions acquire higher kinetic energy and move with higher speeds thus increasing conductivity. Therefore, this option is incorrect

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Ans: Correct Option: (iii)

According to the electrochemical series higher the positive value of standard reduction potential of metal ion, the higher is the oxidizing power

Therefore, MnO₄⁻ is the strongest oxidizing agent.

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Ans: Correct Option: (iii)

According to the electrochemical series, the lower the reduction potential higher is the reducing power.

Therefore, the order of reducing power is Mn²⁺ < Cl– < Cr³⁺ < Cr

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Ans: Correct Option: (iv)

is the highest positive value among the given species. So Mn²⁺ is the most stable ion in its reduced form.

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Ans: Correct Option: (i)

According to the given table, has the most negative value among the given species. So

Cr³⁺ is the most stable oxidized species.

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Correct Option: (iii)

 Ans: Given: One mole of aluminium

Calculate the number of electrons required to convert Al₂O₃ to Al

It is the charge required to obtain one mole of aluminium from Al₂O₃

The dissociation of aluminium oxide is as follows;

Al₂O₃→2Al³⁺+3O²⁻

Al³⁺+3e⁻→Al

Therefore 3Fcharge is required to obtain one mole of aluminum from Al₂O_3.

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Ans: Correct Option: (iv)

The cell constant is described as the ratio of the length of the object and the area of cross-section. It can be written as follow

G= $\frac{\mathrm{l}}{\mathrm{A}}$

As l and A remains constant for an object it can be inferred that the cell constant for a conductivity cell remains constant for a cell.

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Ans: Correct Option: (i)

In the lead storage battery, a reverse reaction occurs and lead sulfate is formed at the anode, and at the cathode, it is converted into lead and lead oxide respectively. The reaction is as follows

At the cathode, the reaction goes this way,

PbSO₄ (s)+2e⁻→Pb (s)+SO₄²⁻ (aq), Reduction process

At the anode, the reaction goes this way,

PbSO₄ (s)+2H₂O→PbO₂ (s)+SO₄²⁻+2H⁺ +2e, oxidation process

The overall reaction is as follows;

2PbSO₄ (s)+2H₂O→Pb (s)+PbO₂ (s)+4H+ (aq)+2SO₄²⁻ (aq)

Therefore, at anode lead sulfate is reduced to lead.

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Ans: Correct Option: (ii)

Given: Salt ammonium hydroxide

Apply Kohlrausch Law 

Kohlrausch law of states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte

Λ⁰_{m (NaOH) =}  Λ⁰_{m (NH4}⁺₎ + Λ⁰_{m (OH}^-₎ + Λ⁰_{m (Na}⁺₎ + Λ⁰_{m (OH}^-₎ - Λ⁰_{m (Na}⁺₎- Λ⁰_{m (Cl}^-₎

Λ⁰_{m (NH4OH)}  = Λ⁰_{m (NH4OH)}  + Λ⁰_{m (NaOH)} – Λ⁰ _(NaCl)

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Ans: Correct Option: (iv)

The electrolysis of aqueous sodium chloride is as follows

NaCl→Na⁺ + Cl⁻

H₂O→H⁺ + OH⁻

At the cathode, the reaction goes this way,

H₂O + e⁻ → $\frac{1}{2}$ H₂ + OH⁻

At the anode, the reaction goes this way,

Cl⁻ (aq)→ $\mathrm{}\frac{1}{2}$ Cl₂ (g)+ e⁻ E? _Cell =1.36V

2H₂O (g)→O₂ (g) + 4H⁺ (aq) + 4e⁻ E? _Cell =1.23V

At the anode, the reaction with a lower E? value will be preferred and oxidation of oxygen is a slow process and requires high voltage so the first reaction will take place.

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Ans: Correct Option: (ii, iv)

The lesser the E? value of the redox couple higher is the reducing power

For Cu²⁺ + 2e⁻→Cu; E? =0.34V

For 2H⁺ + 2e⁻→H₂ E? =0.00V

Since the second redox couple has less standard reduction potential than the first so it can be concluded that the redox couple is a stronger oxidizing agent than H⁺/H₂ and copper cannot displace H₂ from acid.

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Ans: Correct Option: Option (i, iii)

In the electrolysis of sulphuric acid, the following reactions occur

2SO₄²⁻ (aq)→S₂O₈²⁻ (aq) + 2e⁻ E? _Cell =1.96V

2H₂O (l)→O₂ (g)+4H⁺ (aq) + 4e⁻ E? _Cell =1.23V

The reaction will lower the value of E? _Cell is preferred at anode so the second reaction is feasible.

H⁺ + e⁻ → $\frac{1}{2}$ H₂ E? _Cell = 0.00V

At the cathode, reduction of water occurs. Therefore, in dilute sulphuric acid solution, hydrogen will be reduced at the cathode.

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Ans: Correct Option: Option (ii, iii)

At equilibrium, ?G $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

-nFE $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{\mathrm{n}\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

For Daniel cell, n=2

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

At equilibrium, E $°$ = 1.1V

1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$

(i) Since ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  this option is incorrect

(ii) As derived 1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

(iii) As derived _{${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$} so this option is correct

(iv) As ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  _,so this option is incorrect

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

Conductivity is due to the movement of ions in the solution. The conductivity of ions depends on the following factors:

(i) nature of electrolyte added

(ii) size of ion produced

(iii) concentration of electrolyte

(iv) nature of the solvent

(v) temperature

Distance between electrodes does not affect the conductivity of an electrolytic solution.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

 Kohlrausch law of states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

Electrolysis of copper sulfate solution is as follows

CuSO₄ ? Cu²⁺ + SO₄²⁻

H₂O? H⁺ + OH⁻

At the cathode, the reaction goes this way,

Cu²⁺ + 2e− → Cu; E? _Cell = 0.34V

H₂O⁻ → $\frac{1}{2}$ H₂ E? _Cell = 0.00V

At the anode, the reaction goes this way,

2SO₄^2- + 2e−→S₂O₈²⁻ E? _Cell =1.96V

2H₂O→O₂ + 4H⁺ + 4e⁻ E? _Cell =1.23V

The reaction will lower the value of E? _Cell is preferred at anode so the second reaction is feasible.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

In the electrolysis process of CuSO₄ the following reactions occur at the half cell which is as follows:

At the cathode, the reaction goes this way,

Cu²⁺+2e⁻→Cu (s)

At the anode, the reaction goes this way,

Cu (s)→Cu²⁺+2e−

Here copper will be deposited at the cathode and copper will dissolve at the anode.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

Conductance is the reciprocal of resistance and conductivity is the reciprocal of resistivity. It can be written as follows

k = $\frac{1}{\mathbf{?}}$

R= $\mathbf{?}\mathbf{}\frac{1}{\mathbf{A}}$

$\mathbf{?}$ = $\frac{\mathbf{R}\mathbf{A}}{\mathbf{l}}$

k  = $\frac{1}{\frac{\mathbf{R}\mathbf{A}}{\mathbf{l}}}$

k = $\frac{1}{\mathbf{R}}\frac{\mathbf{l}}{\mathbf{A}}$

k = $\frac{1}{\mathbf{R}}$ G*

k = $\frac{{\mathbf{G}}^{\mathbf{*}}}{\mathbf{R}}$

G* is cell constant.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

The conductivity in an ionic solution is due to the ions in the solution. It depends on the following factors;

(a)Temperature: On an increase in temperature, the molar conductivity of ionic solutions increases.

(b)Concentration of Electrolytes: An increase in the concentration of electrolytes decreases the molar conductivity as the number of ions per unit volume decreases.

Therefore, options (i, iii) are correct

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (ii, iii)

In the given representation of the cell, the left side represents the oxidation half cell and the right side of the cell represents the reduction half cell

The electrode at which oxidation occurs is called anode and oxidation of magnesium occurs at anode so magnesium is the anode.

Similarly, the electrode at which the reduction occurs is called the cathode, and the reduction of copper occurs at the cathode.

The cell reaction is as follows

Mg + Cu²⁺→Mg²⁺+ Cu

Empirical formula is C₅H₇N

Empirical mass = 81

Molecular mass = 162

So, molecular formula is C₁₀H₁₄N₂

$•$ Values of principal quantum number, n = 1, 2, 3, .

$•$ Values of azimuthal quantum number,   $\mathcal{l}=0,1,2...........,n-1$

$•$ Number of values of magnetic quantum number are $2\mathcal{l}+1.$  

$•$ Values of spin quantum number are $\pm \frac{1}{2}.$

$•$ Number of orbitals for particular value of $\mathcal{l}are2\mathcal{l}+1,sofor\mathcal{l}=5,$  no. of orbitals = 11.

Geometry of SF₄ is trigonal bipyramidal, in which there is one lonepair which occupy equatorial position as,

There are two lone pair – bond pair repulsions at 90°

pH of acidic buffer is given by,

pH = pKa + log₁₀ $\frac{\left[Salt\right]}{\left[acid\right]}$ $\frac{}{}$

or, $p=pka+lo{g}_{10}\frac{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}{\left[C{H}_{3}C{H}_{2}COOH\right]}$ ${}_{}\frac{{}_{}{}_{}{}^{}}{{}_{}{}_{}}$

or, $\left[H^{+}\right]=k_a\frac{\left[C{H}_{3}C{H}_{2}COOH\right]}{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}$ ${}^{}{}_{}\frac{{}_{}{}_{}}{{}_{}{}_{}{}^{}}$

$\frac{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}{\left[C{H}_{3}C{H}_{2}COOH\right]}=\frac{k_a}{\left[H^{+}\right]}=\frac{1.3\times 10^{-5}}{10^{-4}}=0.13$

Metallic suphides i.e CdS are negatively charged colloidal solution.

Starch colloid is macro molecular colloidal solution.

Fe₂O₃. xH₂O sol is positively charged colloidal solution.

Cheese is an example of gel.

· Cl₂O₇ being non-metallic oxide is acidic in nature.

· Na₂O being metallic oxide is basic in nature.

· Al₂O₃ is an amphoteric oxide

· N₂O is neutral oxide

In extraction of copper from copper pyrite (CuFeS₂).

FeS converted to FeO as;

$\underset{Gangue}{2FeS}+3O_2\stackrel{\Delta }{\to }\underset{Gangue}{2FeO}+2S{O}_2$          

Then FeO is removed as slag by given reaction

FeO + SiO₂ ®   $\underset{Slag}{FeSi{O}_3}$

Isotopes are the atom of same element with different atomic mass.

CaO – Basic Oxide

SO₃ & SiO₂ – Acidic oxides

Photochemical smog occurs in warm, dry and sunny climate. The main components of photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Photochemical smog has high concentration of oxidising agents so called as oxidizing smog.

During nitration using HNO₃ and H₂SO₄, H₂SO₄ act as an acid while HNO₃ act as a base.

Natural rubber is a linear polymer of isoprene called cis-polyisoprene and has elastic properties.

Ci-polyisoprene molecule consists of various chains held together by weak vander Waals interactions and has a coiled structure.

Compound X is sucrose as,

Here, glucose is A which on oxidation with HNO₃ gives saccharic acid as

B is b-D fructose which is laevarotatory.

Using : PV = nRT

1.5 × 416 = n × 0.083 × 300

n = 25mol =  $\frac{W}{M}$

25 =  $\frac{100}{M}$

So, molar mass, M = 4 g/mol.

For combustion of Mg:

Mg (s) + $\frac{1}{2}{O}_2\left(g\right)\to MgO\left(s\right)$

Here,   $\Delta ng=-\frac{1}{2}$

Now using

$\Delta H=\Delta U+\Delta ngRT$

-601.7 =   $\Delta U+\left(-\frac{1}{2}\right)\times 8.3\times 10^{-3}\times 300$

$\Delta U=-600.45kJ$

So; magnitude of   $\Delta U$ is 600 kJ (the nearest integer).

$E_{S{n}^{2+}/Sn}^0=-0.140V=E_2^0$

$E_{S{n}^{4+}/Sn}^0=+0.010V=E_3^0$

Here, $\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$

$-4F{E}_3^0=-2F{E}_1^0-2F{E}_2^0$

$E_3^0=\frac{E_1^0+E_2^0}{2}$

$+0.010=\frac{E_1^0+\left(-0.140\right)}{2}$

$E_1^0=E_{S{n}^{4+}/S{n}^{2+}}^0=+0.16$