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The concentration cell is a type of galvanic cell where both the half-cells consist of the same electrolytes but at different concentrations. The application of the concentration cell takes place along with the electrodes. These cells give a small potential difference while achieving the state of chemical equilibrium.
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Key Takeaways: Concentration cell, electrodes, electrolytes, galvanic cell, half-cells, concentration, electrolytic cell, chemical equilibrium, salt bridge, electrochemical cell, electrons
Concentration Cell
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A concentration cell is an electrolytic cell or an electrochemical cell that consists of two half-cells. Both the half-cells carry the same electrodes but simultaneously they vary in terms of concentration. The more concentrated half-cell is diluted and the half cell of lower concentration has its concentration increased via the transfer of electrons between these two half cells. This is done because the cell as a whole strives to reach a state of chemical equilibrium.
For Example, Consider two solutions of the same electrolyte are connected with a salt bridge. When the two plates of the same metal are dipped separately into the given two solutions, then the whole arrangement is found to act as a galvanic cell.
For a better understanding, follow the diagram illustrated below.
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Types of Concentration Cell
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Concentration Cells are classified into two types. These are-
- Electrode Concentration Cells
- Electrolyte Concentration Cells
Electrode Concentration Cell
- Electrode concentration cells are the ones that consist of two identical solutions. These solutions are used as an electrolyte in each half-cell.
- Although, the half-cells differ in the concentration of the electrode the electrodes are made up of the same material.
- For instance, this type of cell would be a cell that shall consist of two hydrogen electrodes and these electrodes might be varying in terms of pressures but are immersed in the same solutions (containing hydrogen ions).
- The type of cell that will be constituted by the above statement is as follows.
(Pt: H2 (Pressure p1)) / Anode |H+ || (H2 (Pressure p2) Pt) / Cathode Pt: H2(p1) | H+ ions solution | H2(p2): Pt
A tabular illustration of the same is given below.
| R. H. E. | L. H. E. | Overall |
| 2H+ + 2e– → H2(p2) | H2(p1) → 2H+ + 2e– | H2 (p) → H2 (p2) |
- The emf of the cell is given by the following expression
![The emf of the cell is given by the following expression]()
Electrolyte Concentration Cell
- The electrolyte concentration cell mainly consists of identical electrodes which are immersed in the solutions of the same electrolytes.
- The electrolytes of these solutions vary in terms of their concentrations.
- In these cells, the electrolyte tends to diffuse from the solution with the higher concentration towards the solutions of the lower concentration.
- For Example, this type of cell would be a cell that shall consist of an anode that has Zn / Zn2+ (0.1M) whereas the cathode consists of Zn2+ (0.01M) / Zn. Therefore, in this cell, the flow of the electrons from the anode to the cathode is due to the reduction of Zn2+ ions at the cathode into metallic zinc.
- The type of cell that will be constituted by the above statement is as follows.
(Zn|Zn2+ (C1)) / Anode || (Zn2+ (C2 )|Zn) / Cathode
A tabular illustration of the same is given below.
| Anode | Cathode |
| Zn → Zn2+ (C1) + 2e | Zn2+ (C2) + 2e– → Zn (s) |
- The emf of the cell is given by the following expression at 25o C:
![The emf of the cell is given by the following expression at 25o C]()
Components of the Concentration Cell
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Salt Bridge
The salt bridge provides a perfect solution for the separation of the two half - cells and at the same time, it is meant to provide a pathway for ion transfer.
In it, the electric wires would react with the ions which are flowing through them. In that case, the absence of a salt bridge might also lead to a build-up of electrons in the one-half cell from the incoming flow of electrons belonging to the other half cell.
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Electrode
Every Concentration Cell has two electrodes. The two electrodes are called the cathode (right side) and the anode (left side).
The anode tends to lose the electrons and is the site where the oxidation occurs, whereas on the other hand, the cathode tends to gain the electrons, therefore, it is an area where the electrons accumulate and the reduction occurs.
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Voltmeter
In a concentration cell, the voltmeter is used to measure the cell potential of the cell. The cell potential is also referred to as electromotive force / E.M.F. which mainly occurs as a result of the flow of electrons. The voltmeter is generally placed in between the two half-cells.
The value that occurred could be either positive or negative, this depends upon the direction of the flow of the electrons. The potential is then measured in volts or millivolts.
Things to Remember
- A concentration cell is an electrochemical cell where both the half-cells carry the same electrodes but they vary in terms of concentration.
- In a concentration cell, the more concentrated half-cell is diluted and the lower concentration half cell has its concentration increased via the transfer of electrons between these two half cells.
- Concentration Cell is of two types- Electrode concentration cell and electrolyte concentration cell.
- A Concentration cell consists of a salt bridge, electrode and voltmeter.
Also Read:
Ques. Two half cell reactions of an electrochemical cell are given below :
MnO4–(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e–, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours the formation of reactants or products shown in the equation. (All India 2009, 2 Marks)
Ans: The reactions can be represented at the anode and a cathode in the following ways :
At anode (oxidation) :
Sn2+ → Sn4+ (aq) + 2e– ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO4–(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4–(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.
Ques. Calculate the emf of the following cell at 298 K: Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2(g) (1 bar), Pt(s) (Given E°cell = +0.44V) (Delhi 2013, 3 Marks)
Ans: As Fe + 2H+ → Fe2+ + H2 (n = 2)
According to Nernst equation
Ques. Calculate the emf of the following cell at 25°C : Ag(s) | Ag+ (10-3 M) || Cu2+ (10-1 M) | Cu(s) Given that E°cell = +0.46 V and log 10n = n. (All India 2013, 3 Marks)
Ans: Given cell notation is incorrect. Correct cell formula is
Cu2+ (10-1 m) | Cu5 || Ag+ (10-3 M) | Ag(s)
According to Nernst equation
Ecell = 0.31 V
Ques. (a) When a bright silver object is placed in the solution of gold chloride, it acquires a golden tinge but nothing happens when it is placed in a solution of copper chloride. Explain this behaviour of silver.
[Given : E°Cu2+/Cu =+0.34V,E°Ag+/Ag =+0.80V, E°Au3+/Au = +1.40V]
(b) Consider the figure given and answer the following questions :
(i) What is the direction of the flow of electrons?
(ii) Which is the anode and which is the cathode?
(iii) What will happen if the salt bridge is removed?
(iv) How will the concentration of Zn2+ and Ag+ ions be affected when the cell functions?
(v) How will the concentration of these ions be affected when the cell becomes dead? (Comptt. Delhi 2017, 5 Marks)
Ans: (a) The standard electrode potential, E° for silver is 0.80 V and that of gold is 1.5 V. Hence silver can displace gold from its solution. The replaced gold is deposited on silver due to which a golden tinge is obtained. On the other hand, E° for Cu is 0.34 V which is lower than that of silver, thus silver cannot replace copper from its solution.
(b) (i) Electrons flow from anode (Zinc plate) to cathode (Silver plate).
(ii) Zinc plate where oxidation occurs acts as the anode and silver plate where reduction occurs acts as the cathode.
(iii) If the salt bridge is removed then electrons from zinc electrodes will flow to the silver electrode where they will neutralize some of the Ag+ ions and the SO2−4 ions will be left and the solution will acquire a negative charge.
(iv) As silver from silver sulphate solution is deposited on the silver electrode and sulphate ions migrate to the other side, the concentration of AgSO4 solution decreases and of ZnSO4 solution increases as the cell operates.
(v) The concentration of these ions become equal due to the state of equilibrium and zero EMF.
Ques. (a) What is limiting molar conductivity? Why is there a steep rise in the molar conductivity of weak electrolytes on dilution?
(b) Calculate the emf of the following cell at 298 K :
Mg (s) | Mg2+ (0.1 M) || Cu2+ (1.0 × 10-3M) | Cu (s)
[Given = E°Cell = 2.71 V]. (Comptt. Delhi 2017, 5 Marks)
Ans: (a) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ°m. There is a steep rise in the molar conductivity of weak electrolytes on dilution because as the concentration of the weak electrolyte is reduced, there will be an increase in the number of ions in the solution.
(b)
Ques. Determine the flow of electrons for the concentration cell given below.
Fe |Fe2+(0.01 M) || Fe2+(0.1 M) |Fe (6) (6) Fe | Fe2+(0.01 M) || Fe2+ (0.1 M) |Fe (1 mark)
Ans: In the concentration cell as mentioned – above, the electrons shall be transferred from the left side to the right side. This will help the cell in gaining equilibrium. Therefore, as a result, Fe2+ will be formed in the left compartment and metal iron will be deposited on the right electrode.
Ques. A cell consists of two hydrogen electrodes. While the negative electrode is in contact with a solution of 10-6 M hydrogen ions. And the E. M. F. of the cell is 0. 118 volts at 25° Celsius. Calculate the concentration of hydrogen ions at the positive electrode. (2 marks)
Ans: The cell notation can be given as follows:
Pt | H2 (1 atm.) |H+||H+|H2(1 atm.) |Pt
10-6 M CM
Anode: H2 → 2H+ + 2e-
Cathode: 2H+ + 2e → H2
Ecell = 0.0591/2 log ([H+]cathode) / [10-6]2
0.081 = (0. 0591) log ([H+]) / 10-6
Log [H+] cathode / 10-6 =0. 118 / 0. 0591 = 2
[H+] cathode / 10-6 = 102
[H+] cathode = 10-6 = 10-4 M
Ques. Calculate the cell potential for a concentration cell that consists of two silver electrodes with concentrations of 0.2 M and 3.0 M. (1 mark)
Ans: Reaction: Ag2++2e−?Ag(s)
Cell representation: Ag (s) |Ag2+ (0.2 M) || Ag2+ (3.0 M) |Ag (s)
Nernst Equation: E = E0 − 0.0592 / 2 log 0.02 / 3.0
E = 0.0644 V
Ques. A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured at understanding the0,422 V. Determine the concentration of silver ions in the cell.
Given : E°Ag+/Ag = + 0.80 V, E° Cu2+/Cu = +0.34 V. (All India 2009, 3 Marks)
Ans: The reaction takes place at anode and cathode in the following ways :
At anode (oxidation) :
Cu(s) → Cu2+(aq) + 2e–
At cathode (reduction) :
Cu(s) + 2Ag2+(aq) → Cu2+(aq) + 2Ag(s)
The complete cell reaction is
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