Question 11 Mark
Suggest a list of metals that are extracted electrolytically.
Answer
View full question & answer→Na, Ca, Mg and Al
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Question 21 Mark
How much electricity is required in coulomb for the oxidation of
1 mol of FeO to Fe2O3.
1 mol of FeO to Fe2O3.
Answer
View full question & answer→According to the question,
Fe2+ → Fe3+ + e-1
Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F
= 96487 C
Fe2+ → Fe3+ + e-1
Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F
= 96487 C
Question 31 Mark
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Al, Cu, Fe, Mg and Zn.
Answer
View full question & answer→Mg > Al > Zn > Fe > Cu is decreasing order of their reactivity.
Question 41 Mark
Why does the conductivity of a solution decrease with dilution?
Answer
View full question & answer→The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.
Question 51 Mark
How much electricity is required in coulomb for the oxidation of
1 mol of H2O to O2.
1 mol of H2O to O2.
Answer
View full question & answer→According to the question,
$\text{H}_2\text{O}\ \rightarrow\ \text{H}_2\ +\ \frac{1}{2}\text{O}_2$
Now, we can write:
$\text{O}_2\rightarrow\frac{1}{2}\text{O}_2+2\text{e}^-$
Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F
= 2 × 96487 C
= 192974 C
$\text{H}_2\text{O}\ \rightarrow\ \text{H}_2\ +\ \frac{1}{2}\text{O}_2$
Now, we can write:
$\text{O}_2\rightarrow\frac{1}{2}\text{O}_2+2\text{e}^-$
Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F
= 2 × 96487 C
= 192974 C
Question 61 Mark
How much charge is required for the following reduction:
1 mol of Al3+ to Al.
1 mol of Al3+ to Al.
Answer
View full question & answer→Al3+ + 3e- → Al
Therefore, Required charge = 3 F
= 3 × 96487 C
= 289461 C
Therefore, Required charge = 3 F
= 3 × 96487 C
= 289461 C
Question 71 Mark
How much charge is required for the following reduction:
1 mol of Cu2+ to Cu.
1 mol of Cu2+ to Cu.
Answer
View full question & answer→Cu2+ + 2e- → Cu
Therefore, Required charge = 2 F
= 3 × 96487 C
= 192974 C
Therefore, Required charge = 2 F
= 3 × 96487 C
= 192974 C
Question 81 Mark
Given the standard electrode potentials,
K+/K = –2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74V
Arrange these metals in their increasing order of reducing power.
K+/K = –2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74V
Arrange these metals in their increasing order of reducing power.
Answer
View full question & answer→Ag < Hg < Cr < Mg < K.
Question 91 Mark
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer
View full question & answer→Methane and Methanol.
Question 101 Mark
How much electricity in terms of Faraday is required to produce,
40.0g of Al from molten Al2O3.
40.0g of Al from molten Al2O3.
Answer
View full question & answer→According to the question,
$\text{Al}^{3+}+3\text{e}^{-1}\ \rightarrow\ \text{Al}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 27\text{g}$
Electricity required to produce 27g of Al = 3 F
Therefore, electricity required to produce 40 g of Al $=\frac{3\times40}{27}\text{F}=4.44\ \text{F}$
$\text{Al}^{3+}+3\text{e}^{-1}\ \rightarrow\ \text{Al}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 27\text{g}$
Electricity required to produce 27g of Al = 3 F
Therefore, electricity required to produce 40 g of Al $=\frac{3\times40}{27}\text{F}=4.44\ \text{F}$
Question 111 Mark
Can you store copper sulphate solutions in a zinc pot?
Answer
View full question & answer→Zn being more reactive than Cu, displaces Cu from CuSO4 solution as follows: Zn (s) + CuSO4 (aq) → ZnSO4(ag) + Cu (s)
In terms of EMF, we have
Zn|Zn2+||Cu2+|Cu
$\text{E}^\circ_\text{cell}=\text{E}^\circ_{\text{Cu}^{2+}\text{Cu}}-\text{E}^\circ_{\text{Zn}^{2+}/\text{Zn}}$
= 0.34 V - (-0.76 V)
= 1.10 V
As $\text{E}^\circ_\text{cell}$ is positive, reaction takes place, i.e., Zn reacts with copper and hence, we cannot store CuSo4 solution in zinc pot.
In terms of EMF, we have
Zn|Zn2+||Cu2+|Cu
$\text{E}^\circ_\text{cell}=\text{E}^\circ_{\text{Cu}^{2+}\text{Cu}}-\text{E}^\circ_{\text{Zn}^{2+}/\text{Zn}}$
= 0.34 V - (-0.76 V)
= 1.10 V
As $\text{E}^\circ_\text{cell}$ is positive, reaction takes place, i.e., Zn reacts with copper and hence, we cannot store CuSo4 solution in zinc pot.
Question 121 Mark
Consider the reaction:
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 8H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72–?
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 8H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72–?
Answer
View full question & answer→Cr2O72– → 2Cr3+
2Cr6+ + 6e– → 2Cr3+
In given equation there are 6 electrons are required so that n = 6
Use the formula
Required charge = nF
Plug the values in this formula we get
Required charge = 6 × 96487 Coulombs
= 578922 Coulombs
= 5.79 × 105 Coulombs
2Cr6+ + 6e– → 2Cr3+
In given equation there are 6 electrons are required so that n = 6
Use the formula
Required charge = nF
Plug the values in this formula we get
Required charge = 6 × 96487 Coulombs
= 578922 Coulombs
= 5.79 × 105 Coulombs
Question 131 Mark
How much charge is required for the following reduction:
1 mol of MnO4– to Mn2+.
1 mol of MnO4– to Mn2+.
Answer
View full question & answer→MnO-4 → Mn2+
i.e., Mn7+ + 5e- → Mn2+
Therefore, Required charge = 5 F
= 5 × 96487 C
= 482435 C
i.e., Mn7+ + 5e- → Mn2+
Therefore, Required charge = 5 F
= 5 × 96487 C
= 482435 C
Question 141 Mark
How much electricity in terms of Faraday is required to produce,
20.0 g of Ca from molten CaCl2.
20.0 g of Ca from molten CaCl2.
Answer
View full question & answer→According to the question, $\text{Ca}^{2+}+2\text{e}^{-1}\ \rightarrow\ \text{Ca}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 40\text{g}$ Electricity required to produce 40 g of calcium = 2 F Therefore, electricity required to produce 20 g of calcium $\frac{2\times20}{40}\text{F}=1\text{F}$
Question 151 Mark
What is meant by ‘limiting molar conductivity’?
Answer
View full question & answer→It is molar conductivity at infinite dilution or approaching zero concentration.
Question 161 Mark
Express the relation between the conductivity and the molar conductivity of a solution.
Answer
View full question & answer→$\Lambda_{m}=\frac{k}{\text{c}}$ where Λm is molar conductivity, κ is conductivity, c is concentration in mol L-1.
Question 171 Mark
From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following:
Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following:
- Which cell is used in hearing aids?
- Which cell was used in Apollo Space Programme?
- Which cell is used in automobiles and inverters?
- Which cell does not have long life?
Answer
View full question & answer→- Mercury cell.
- Fuel cell.
- Lead storage battery.
- Dry cell.
Question 181 Mark
Express the relation between conductivity and molar conductivity of a solution held in a cell.
Answer
View full question & answer→$\Lambda$m = k/c.
where $\Lambda$m is molar conductivity, k is conductivity, c is concentration in mol L–1.
where $\Lambda$m is molar conductivity, k is conductivity, c is concentration in mol L–1.
Question 191 Mark
Out of zinc and tin, whose coating is better to protect iron objects?
Answer
View full question & answer→Zink is better than tin in protecting iron from corrosion because zink has more affinity to oxygen than tin. When is coated on iron layer then it reacts with oxygen if air to form a protective layer of zink oxide on iron which prevent the further reaction of iron with oxygen, and thus preventing the process of rusting. But in the case of tin, it(tin) does not react with oxygen thus it is less effective in preventing rusting.
Question 201 Mark
Write Nernst equation for the reaction
$2\text{Cr}+3\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ }2\text{Cr}^{3+}+3\text{Fe}$
$2\text{Cr}+3\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ }2\text{Cr}^{3+}+3\text{Fe}$
Answer
View full question & answer→$\text{E}_{\text{cell}}=\text{E}^{\circ}_{\text{cell}}-\frac{\text{RT}}{6\text{F}}\text{In}\frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3}$ $(\because\text{n}=6)$
Question 211 Mark
Consider a cell given below $\text{Cu}|\text{Cu}^{2+}||\text{Cl}^{-}|\text{Cl}_2,\text{pt}$ Write the reactions that occur at anode and cathode.
Answer
View full question & answer→$\text{Anode}:\ \ \text{Cu}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{Cu}^{2+}+2\text{e}^{-}$
$\text{Cathode}:\ \ \text{Cl}_2+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{Cl}^{-}\text{Cu}$ is anode as it is getting oxidised.
Cl2 is cathode as it is getting reduced.
$\text{Cathode}:\ \ \text{Cl}_2+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{Cl}^{-}\text{Cu}$ is anode as it is getting oxidised.
Cl2 is cathode as it is getting reduced.
Question 221 Mark
What is the role of ZnCl2 in a dry cell?
Answer
View full question & answer→ZnCl2 combines with the NH3 produced to form the complex salt [Zn(NH3)2Cl2] otherwise the pressure developed due to NH3 would crack the seal of the cell.
Question 231 Mark
What are secondary cells?
Answer
View full question & answer→Secondary cells are those cells which are rechargeable, i.e., the products can be changed back to reactants.
Question 241 Mark
What is meant by cell constant?
Answer
View full question & answer→Cell constant is the ratio of distance (l) between electrodes and area of cross-section (A). It is denoted by $\frac{\text{l}}{\text{A}}.$ Its unit is cm-1.
Question 251 Mark
What is galvanization?
Answer
View full question & answer→The process of coating zinc over iron is called galvanization.
Question 261 Mark
Value of standard electrode potential for the oxidation of Cl- ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl- oxidised at anode instead of water?
Answer
View full question & answer→Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires overpotential hence Cl- is oxidised instead of water.
Question 271 Mark
Why does a galvanic cell become dead after some time?
Answer
View full question & answer→As the reaction proceeds, concentration of ions in anodic half keeps on increasing while in the cathodic half it keeps on decreasing. Hence, their electrode potentials also keep on changing till ultimately they become equal and then e.m.f. of the cell becomes zero.
Question 281 Mark
Three iron sheets have been coated separately with three metals A, B and C whose standard electrode potentials are given below.
Identify in which case rusting will take place faster when coating is damaged.
| Metal | A | B | C | Iron |
| E° | -0.46V | -0.66V | -0.20V | -0.44V |
Answer
View full question & answer→As iron (-0.44V) has lower standard reduction potential than C (-0.20V) only therefore when coating is broken, rusting will take place faster.
Question 291 Mark
Why is alternating current used for measuring resistance of an electrolytic solution?
Answer
View full question & answer→Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant.
Question 301 Mark
Using the E° values of A and B, predict which is better for coating the surface of iron $\Big[\text{E}^{\circ}_{\text{Fe}^{2+}/\text{Fe}}=-0.44\text{V}\Big]$ to prevent corrosion and why?
Given: $\Big[\text{E}^{\circ}_{\text{A}^{2+}/\text{A}}=-2.37\text{V, E}^{\circ}_{\text{B}^{2+}/\text{B}}=-0.14\text{V}\Big]$
Given: $\Big[\text{E}^{\circ}_{\text{A}^{2+}/\text{A}}=-2.37\text{V, E}^{\circ}_{\text{B}^{2+}/\text{B}}=-0.14\text{V}\Big]$
Answer
View full question & answer→A, as its standard reduction potential is less than B, therefore, it will undergo oxidation more easily than B.
Question 311 Mark
Depict the galvanic cell in which the cell reaction is $\text{Cu}+2\text{Ag}^{+}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }2\text{Ag}+\text{Cu}^{2+}$
Answer
View full question & answer→$\text{Cu}|\text{Cu}^{2+}||\text{Ag}^{+}|\text{Ag}$
Question 321 Mark
What is electrode potential?
Answer
View full question & answer→The electrical potential difference set up between the metal and its solution is called electrode potential.
Question 331 Mark
What is the overall electrochemical reaction taking place in rusting?
Answer
View full question & answer→$2\text{Fe(s) + O}_2(\text{g})+4\text{H}^+(\text{aq})\xrightarrow{ \ \ \ \ \ \ }2\text{Fe}^{2+}(\text{aq})+2\text{H}_2\text{O}(\text{I})$
Question 341 Mark
Define specific conductance or conductivity.
Answer
View full question & answer→Specific conductance is defined as conductance of electrolyte when distance between electrodes is 1cm and area of cross section is 1cm2.
Question 351 Mark
Why electrolysis of NaBr and NaI gives Br2 and I2 respectively while that of NaF gives O2 instead of F2?
Answer
View full question & answer→Br- and I- ions have higher oxidation potentials than water. Hence, they are more easily oxidised. But F- ions have lower oxidation potential than H2O. So, H2O is easily oxidised to give O2 gas.
Question 361 Mark
Why is it not possible to measure the single electrode potential?
Answer
View full question & answer→Oxidation or reduction cannot take place alone. Moreover, electrode potential is a relative tendency and can be measured with respect to a reference electrode only.
Question 371 Mark
What is the reference electrode in determining the standard electrode potential?
Answer
View full question & answer→Normal hydrogen electrode (NHE).
Question 381 Mark
In an aqueous solution how does specific conductivity of electrolytes change with addition of water?
Answer
View full question & answer→On the addition of water, number of ions per unit volume decreases and therefore conductivity decreases.
Question 391 Mark
Write Nernst equation for the general cell reaction,
$\text{aA}+\text{bB}\xrightarrow{ \ \ \ \ \ \ }\text{xX + yY}.$
$\text{aA}+\text{bB}\xrightarrow{ \ \ \ \ \ \ }\text{xX + yY}.$
Answer
View full question & answer→If n moles of electrons are transferred, Nernst equation is:
$\text{E}_{\text{cell}}=\text{E}^{\circ}_{\text{cell}}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{[\text{X}]^{\text{z}}[\text{Y}]^{\text{y}}}{[\text{A}]^{\text{a}}[\text{B}]^{\text{b}}}$
$\text{E}_{\text{cell}}=\text{E}^{\circ}_{\text{cell}}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{[\text{X}]^{\text{z}}[\text{Y}]^{\text{y}}}{[\text{A}]^{\text{a}}[\text{B}]^{\text{b}}}$
Question 401 Mark
Which type of a metal can be used in cathodic protection of iron against rusting?
Answer
View full question & answer→A metal which is more electropositive than iron such as Al, Zn, Mg can be used in cathodic protection of iron against rusting.
Question 411 Mark
What are the products of electrolysis of molten and aqueous sodium chloride?
Answer
View full question & answer→- Na and Cl2.
- H2 and Cl2.
Question 421 Mark
Write the name of the electrolyte used in (i) fuel cell (ii) mercury cell.
Answer
View full question & answer→- Concentrated aqueous KOH solution.
- Moist mercuric oxide (HgO) mixed with KOH.
Question 431 Mark
How can the reduction potential of an electrode be increased?
Answer
View full question & answer→$\text{M}^{\text{n}+}+\text{ne}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{M},$
$\frac{\text{E}_{\text{M}^{\text{n}+}}}{\text{M}}=\frac{\text{E}^{\circ}_{\text{M}^{\text{n}+}}}{\text{M}}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{1}{[\text{M}^{\text{n}+}]}$
$=\frac{\text{E}^{\circ}_{\text{M}^{\text{n}+}}}{\text{M}}+\frac{\text{RT}}{\text{nF}}\text{In}[\text{M}^{\text{n}+}]$
Thus, electrode potential can be increased by increasing the metal ion concentration.
$\frac{\text{E}_{\text{M}^{\text{n}+}}}{\text{M}}=\frac{\text{E}^{\circ}_{\text{M}^{\text{n}+}}}{\text{M}}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{1}{[\text{M}^{\text{n}+}]}$
$=\frac{\text{E}^{\circ}_{\text{M}^{\text{n}+}}}{\text{M}}+\frac{\text{RT}}{\text{nF}}\text{In}[\text{M}^{\text{n}+}]$
Thus, electrode potential can be increased by increasing the metal ion concentration.
Question 441 Mark
Why does an aqueous solution of NaCl on electrolysis give H2 gas at the cathode and not sodium metal?
Answer
View full question & answer→This is because of the fact that standard reduction potential of water is greater than that of sodium.
Question 451 Mark
What is the necessity to use a salt bridge in a Galvanic cell?
Answer
View full question & answer→To complete the inner circuit and to maintain the electrical neutrality of the electrolytic solutions of the half-cells we use a salt bridge in a Galvanic cell.
Question 461 Mark
State the factors that influence the value of cell potential of the following cell:
Mg(s) | Mg2+(aq) || Ag+(aq) | Ag(s)
Mg(s) | Mg2+(aq) || Ag+(aq) | Ag(s)
Answer
View full question & answer→Concentration of Mg 2+ and Ag+ ions in the solution and temperature.
Question 471 Mark
What flows in the internal circuit of a galvanic cell?
Answer
View full question & answer→Ions.
Question 481 Mark
Write the chemical formula of rust.
Answer
View full question & answer→Fe2O3. xH2O
Question 491 Mark
Which of the statements about solutions of electrolytes is not correct?
- Conductivity of solution depends upon size of ions.
- Conductivity depends upon viscosiy of solution.
- Conductivity does not depend upon solvation of ions present in solution.
- Conductivity of solution increases with temperature.
Answer
View full question & answer→- Conductivity does not depend upon solvation of ions present in solution.
Explanation:
Conductivity depends upon salvation of ions present in the solution. The greater the salvation of ions, the lesser is the conductivity.
Question 501 Mark
View full question & answer→
