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Q2.7:
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Solution :
1. Definitions of Conductivity and Molar Conductivity
Conductivity ($\kappa$): Conductivity, or specific conductance, is defined as the conductance of a solution of an electrolyte contained between two electrodes of unit area of cross-section separated by a unit distance. Mathematically, if $R$ is the resistance, $l$ is the length, and $A$ is the area of cross-section, then $R = \rho \frac{l}{A}$. Since conductivity ($\kappa$) is the reciprocal of resistivity ($\rho$), we have:
$\kappa = \frac{1}{R} \times \frac{l}{A} = G \times G^*$
Where $G$ is conductance and $G^*$ is the cell constant. The SI unit of conductivity is $\text{S m}^{-1}$.
Molar Conductivity ($\Lambda_m$): Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution of volume $V$. It is related to conductivity by the expression:
$\Lambda_m = \frac{\kappa \times 1000}{M}$
Where $M$ is the molarity of the solution. The SI unit is $\text{S m}^2 \text{mol}^{-1}$.
2. Variation with Concentration
The variation of these parameters with concentration ($C$) is dictated by the availability of charge carriers and their mobility.
| Parameter | Variation with Dilution (Decrease in Concentration) | Theoretical Justification |
|---|---|---|
| Conductivity ($\kappa$) | Decreases | [Per the Arrhenius theory], $\kappa$ depends on the number of ions per unit volume. Upon dilution, the number of ions per unit volume decreases, leading to a decrease in $\kappa$. |
| Molar Conductivity ($\Lambda_m$) | Increases | Although $\kappa$ decreases, the volume ($V$) containing one mole of electrolyte increases significantly. The increase in volume outweighs the decrease in $\kappa$. |
3. Graphical Representation of Molar Conductivity
4. Analytical Discussion
Strong Electrolytes: [Per the Kohlrausch law and Debye-Hückel-Onsager equation], $\Lambda_m$ increases linearly with $\sqrt{C}$ according to $\Lambda_m = \Lambda_m^\circ - A\sqrt{C}$. As concentration approaches zero, $\Lambda_m$ reaches a limiting value called molar conductivity at infinite dilution ($\Lambda_m^\circ$).
Weak Electrolytes: These show a sharp increase in $\Lambda_m$ at low concentrations. This is due to the increase in the degree of dissociation ($\alpha$) as dilution increases [Per Ostwald’s Dilution Law]. $\Lambda_m$ approaches infinity as $C \rightarrow 0$ because the degree of dissociation approaches unity.
Final Solution: Conductivity decreases with dilution due to reduced ion density per unit volume, while molar conductivity increases with dilution due to increased dissociation and volume expansion per mole.
More Questions from Class 12 Chemistry Electrochemistry Exercises
- Q2.1: 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.
- Q2.1: 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.
- Q2.10: The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Calculate (\Lambda_{m}) for all concentrations and draw a plot between (\Lambda_{m}) and (c^{1/2}). Find the value of (\Lambda^{0}_{m}).
- Q2.10: The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Calculate \(\Lambda_{m}\) for all concentrations and draw a plot between \(\Lambda_{m}\) and \(c^{1/2}\). Find the value of \(\Lambda_{m}^{\circ}\).
- Q2.11: Conductivity of 0.00241 M acetic acid is 7.896 (\times) (10^{-5}) S (cm^{-1}). Calculate its molar conductivity. If (\Lambda^{0}_{m}) for acetic acid is 390.5 S (cm^{2} mol^{-1}), what is its dissociation constant?
- Q2.11: Conductivity of 0.00241 M acetic acid is \(7.896 \times 10^{-5} S cm^{-1}\). Calculate its molar conductivity. If \(\Lambda_{m}^{\circ}\) for acetic acid is \(390.5 S cm^{2} mol^{-1}\), what is its dissociation constant?
- Q2.12 (i): How much charge is required for the following reduction: (i) 1 mol of (Al^{3+}) to Al?
- Q2.12 (ii): How much charge is required for the following reduction: (ii) 1 mol of (Cu^{2+}) to Cu?
- Q2.12 (iii): How much charge is required for the following reduction: (iii) 1 mol of (MnO_{4}^{-}) to (Mn^{2+})?
- Q2.12(i): How much charge is required for the following reduction: 1 mol of \(Al^{3+}\) to Al?
- Q2.12(ii): How much charge is required for the following reduction: 1 mol of \(Cu^{2+}\) to Cu?
- Q2.12(iii): How much charge is required for the following reduction: 1 mol of \(MnO_{4}^{-}\) to \(Mn^{2+}\)?
- Q2.13 (i): How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten (CaCl_{2})?
- Q2.13 (ii): How much electricity in terms of Faraday is required to produce (ii) 40.0 g of Al from molten (Al_{2}O_{3})?
- Q2.13(i): How much electricity in terms of Faraday is required to produce 20.0 g of Ca from molten \(CaCl_{2}\)?
- Q2.13(ii): How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten \(Al_{2}O_{3}\)?
- Q2.14 (i): How much electricity is required in coulomb for the oxidation of (i) 1 mol of (H_{2}O) to (O_{2})?
- Q2.14 (ii): How much electricity is required in coulomb for the oxidation of (ii) 1 mol of (FeO) to (Fe_{2}O_{3})?
- Q2.14(i): How much electricity is required in coulomb for the oxidation of 1 mol of \(H_{2}O\) to \(O_{2}\)?
- Q2.14(ii): How much electricity is required in coulomb for the oxidation of 1 mol of FeO to \(Fe_{2}O_{3}\)?
- Q2.15: 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?
- Q2.15: 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?
- Q2.16: Three electrolytic cells A,B,C containing solutions of (ZnSO_{4}), (AgNO_{3}) and (CuSO_{4}), 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?
- Q2.16: Three electrolytic cells A,B,C containing solutions of \(ZnSO_{4}\), \(AgNO_{3}\) and \(CuSO_{4}\), 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?
- Q2.17 (i): Using the standard electrode potentials, predict if the reaction between the following is feasible: (i) (Fe^{3+}(aq)) and (I^{-}(aq))
- Q2.17 (ii): Using the standard electrode potentials, predict if the reaction between the following is feasible: (ii) (Ag^{+} (aq)) and Cu(s)
- Q2.17 (iii): Using the standard electrode potentials, predict if the reaction between the following is feasible: (iii) (Fe^{3+} (aq)) and (Br^{-} (aq))
- Q2.17 (iv): Using the standard electrode potentials, predict if the reaction between the following is feasible: (iv) Ag(s) and (Fe^{3+} (aq))
- Q2.17 (v): Using the standard electrode potentials, predict if the reaction between the following is feasible: (v) (Br_{2} (aq)) and (Fe^{2+} (aq)).
- Q2.17(i): Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: \(Fe^{3+}(aq)\) and \(I^{-}(aq)\)
- Q2.17(ii): Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: \(Ag^{+} (aq)\) and \(Cu(s)\)
- Q2.17(iii): Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: \(Fe^{3+} (aq)\) and \(Br^{-} (aq)\)
- Q2.17(iv): Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: \(Ag(s)\) and \(Fe^{3+} (aq)\)
- Q2.17(v): Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: \(Br_{2} (aq)\) and \(Fe^{2+} (aq)\).
- Q2.18 (i): Predict the products of electrolysis in each of the following: (i) An aqueous solution of (AgNO_{3}) with silver electrodes.
- Q2.18 (ii): Predict the products of electrolysis in each of the following: (ii) An aqueous solution of (AgNO_{3}) with platinum electrodes.
- Q2.18 (iii): Predict the products of electrolysis in each of the following: (iii) A dilute solution of (H_{2}SO_{4}) with platinum electrodes.
- Q2.18 (iv): Predict the products of electrolysis in each of the following: (iv) An aqueous solution of (CuCl_{2}) with platinum electrodes.
- Q2.18(i): Predict the products of electrolysis in each of the following: An aqueous solution of \(AgNO_{3}\) with silver electrodes.
- Q2.18(ii): Predict the products of electrolysis in each of the following: An aqueous solution of \(AgNO_{3}\) with platinum electrodes.
- Q2.18(iii): Predict the products of electrolysis in each of the following: A dilute solution of \(H_{2}SO_{4}\) with platinum electrodes.
- Q2.18(iv): Predict the products of electrolysis in each of the following: An aqueous solution of \(CuCl_{2}\) with platinum electrodes.
- Q2.2: Given the standard electrode potentials, (K^{+}/K) = –2.93V, (Ag^{+}/Ag) = 0.80V, (Hg^{2+}/Hg) = 0.79V (Mg^{2+}/Mg) = –2.37 V, (Cr^{3+}/Cr) = – 0.74V Arrange these metals in their increasing order of reducing power.
- Q2.2: Given the standard electrode potentials, \(K^{+}/K = –2.93V\), \(Ag^{+}/Ag = 0.80V\), \(Hg^{2+}/Hg = 0.79V\) \(Mg^{2+}/Mg = –2.37 V\), \(Cr^{3+}/Cr = – 0.74V\) Arrange these metals in their increasing order of reducing power.
- Q2.3 (i): Depict the galvanic cell in which the reaction (Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)) takes place. Further show: (i) Which of the electrode is negatively charged?
- Q2.3 (ii): Depict the galvanic cell in which the reaction (Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)) takes place. Further show: (ii) The carriers of the current in the cell.
- Q2.3 (iii): Depict the galvanic cell in which the reaction (Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)) takes place. Further show: (iii) Individual reaction at each electrode.
- Q2.3(i): Depict the galvanic cell in which the reaction \(Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)\) takes place. Further show: Which of the electrode is negatively charged?
- Q2.3(ii): Depict the galvanic cell in which the reaction \(Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)\) takes place. Further show: The carriers of the current in the cell.
- Q2.3(iii): Depict the galvanic cell in which the reaction \(Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)\) takes place. Further show: Individual reaction at each electrode.
- Q2.4 (i): Calculate the standard cell potentials of galvanic cell in which the following reaction takes place: (i) (2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd) Calculate the (\Delta_{r}G^{o}) and equilibrium constant of the reaction.
- Q2.4 (ii): Calculate the standard cell potentials of galvanic cell in which the following reaction takes place: (ii) (Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)) Calculate the (\Delta_{r}G^{o}) and equilibrium constant of the reaction.
- Q2.4(i): Calculate the standard cell potentials of galvanic cell in which the following reaction takes place: \(2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)\) Calculate the \(\Delta_{r}G^{\circ}\) and equilibrium constant of the reaction.
- Q2.4(ii): Calculate the standard cell potentials of galvanic cell in which the following reaction takes place: \(Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)\) Calculate the \(\Delta_{r}G^{\circ}\) and equilibrium constant of the reaction.
- Q2.5 (i): Write the Nernst equation and emf of the following cell at 298 K: (i) (Mg(s)|Mg^{2+}(0.001M)||Cu^{2+}(0.0001 M)|Cu(s))
- Q2.5 (ii): Write the Nernst equation and emf of the following cell at 298 K: (ii) (Fe(s)|Fe^{2+}(0.001M)||H^{+}(1M)|H_{2}(g)(1bar)| Pt(s))
- Q2.5 (iii): Write the Nernst equation and emf of the following cell at 298 K: (iii) (Sn(s)|Sn^{2+}(0.050 M)||H^{+}(0.020 M)|H_{2}(g) (1 bar)|Pt(s))
- Q2.5 (iv): Write the Nernst equation and emf of the following cell at 298 K: (iv) (Pt(s)|Br^{-}(0.010 M)|Br_{2}(l )||H^{+}(0.030 M)| H_{2}(g) (1 bar)|Pt(s)).
- Q2.5(i): Write the Nernst equation and emf of the following cell at 298 K: \(Mg(s)|Mg^{2+}(0.001M)||Cu^{2+}(0.0001 M)|Cu(s)\)
- Q2.5(ii): Write the Nernst equation and emf of the following cell at 298 K: \(Fe(s)|Fe^{2+}(0.001M)||H^{+}(1M)|H_{2}(g)(1bar)| Pt(s)\)
- Q2.5(iii): Write the Nernst equation and emf of the following cell at 298 K: \(Sn(s)|Sn^{2+}(0.050 M)||H^{+}(0.020 M)|H_{2}(g) (1 bar)|Pt(s)\)
- Q2.5(iv): Write the Nernst equation and emf of the following cell at 298 K: \(Pt(s)|Br^{-}(0.010 M)|Br_{2}(l )||H^{+}(0.030 M)| H_{2}(g) (1 bar)|Pt(s)\).
- Q2.6: In the button cells widely used in watches and other devices the following reaction takes place: (Zn(s) + Ag_{2}O(s) + H_{2}O(l ) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)) Determine (\Delta_{r}G^{o}) and (E^{o}) for the reaction.
- Q2.6: In the button cells widely used in watches and other devices the following reaction takes place: \(Zn(s) + Ag_{2}O(s) + H_{2}O(l ) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)\) Determine \(\Delta_{r}G^{\circ}\) and \(E^{\circ}\) for the reaction.
- Q2.7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
- Q2.8: The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S (cm^{-1}). Calculate its molar conductivity.
- Q2.8: The conductivity of 0.20 M solution of KCl at 298 K is \(0.0248 S cm^{-1}\). Calculate its molar conductivity.
- Q2.9: The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 (\Omega). What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 (\times) (10^{-3}) S (cm^{-1}).
- Q2.9: The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is \(1500 \Omega\). What is the cell constant if conductivity of 0.001M KCl solution at 298 K is \(0.146 \times 10^{-3} S cm^{-1}\).
CBSE Solutions for Class 12 Chemistry Electrochemistry
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