Cell Potential: Definition, Diagram, Reactions and Difference

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Jasmine Grover

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The difference in potential between two half cells in an electrochemical cell is called the cell potential, Ecell. The potential difference is caused by the ability of electrons to flow from one-half of the cell to the other. Electrons can move between electrodes because the chemical reaction is a redox reaction. When one material is oxidised and another is reduced, this is known as a redox reaction. When a substance is oxidised, it loses one or more electrons and becomes positively charged. During reduction, on the other hand, the substance loses electrons and becomes negatively charged. 

This pertains to the measurement of the cell potential because the cell potential is determined by the difference between the potential for the reducing agent to get oxidised and the potential for the oxidising agent to become reduced. The cell potential (Ecell) is measured in voltage (V), allowing us to assign a numerical value to it.

Key Takeaways: Electrochemical cell, Cell potential, Anode, Cathode, Electron, Voltage, Electricity, Electrode, Oxidising agent, Half cell


Electrochemical Cell

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Chemical reactions, such as the batteries in your remote and the motor in your car, produce electricity in a variety of ways through the passage of electrons. In a battery, the cell potential is used to determine the voltage difference between two half cells. We'll go over how to determine the voltage in an electrochemical cell and what components are involved.

All of the components are listed below:

  • There are two half cells.
  • There are two metal electrodes.
  • A single voltmeter
  • There is one salt bridge.
  • Each half cell requires two aqueous solutions.

The Electrochemical Cell is made up of all of these parts.
Two half cells make up an electrochemical cell. The oxidation of a metal electrode happens in one of the half cells, while the reduction of metal ions in solution occurs in the other half cell. A metal electrode of a certain metal is submerged in an aqueous solution containing the same metal ions in the half cell. The electrode is connected to the other half cell, which has a metal electrode submerged in an aqueous solution of metal ions. The anode will be designated as the first half cell in this situation. The metal atoms in the electrode become oxidised in this half cell and join the other metal ions in the aqueous solution. 

  • A copper electrode is an example of this since the Cu atoms in the electrode lose two electrons and become Cu2+.

The Cu2+ ions would then be added to an aqueous solution that already contained a particular amount of Cu2+ ions. The electrons lost by Cu atoms in the electrode are subsequently transferred to the cathode, the second half cell. In this example, we'll use a silver electrode in an aqueous solution of silver ions as the second half cell. 

The Ag+ ions in the solution will be reduced as electrons are transported to the Ag electrode, resulting in an Ag atom on the Ag electrode. A salt bridge connects the half cells in order to balance the charge on both sides of the cell.

As the Cu2+ ions overwhelm the anode half cell, the salt's negative anion enters the solution and stabilises the charge. Similarly, when the solution gets increasingly negatively charged in the cathode half cell, cations from the salt bridge will stabilise the charge.

  • The electrode potential explains an electrode's tendency to lose or absorb electrons when it comes into contact with an electrolyte. These potentials' values can be utilised to forecast the cell's overall potential. The electrode potentials are usually measured using a conventional hydrogen electrode as a reference electrode (an electrode of known potential).
Electrochemical Cell
Electrochemical Cell

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Cell Diagram 

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The cell diagram shows how the electrochemical cell's overall reaction works. The chemicals that are involved in the reduction and oxidation reactions are the ones that are actually reacting. (The spectator ions aren't included.) The anode half cell is always written on the left side of the cell design, whereas the cathode half cell is usually written on the right side of the cell diagram. Two vertical lines separate the anode from the cathode. 

A single vertical line separates the electrodes of the anode and cathode solutions. If the aqueous solution contains other chemicals, they are added to the graphic with a comma followed by the chemical. If copper wasn't oxidised alone in the image above, and another chemical like K was involved, it would be denoted as (Cu, K) in the diagram. It's easier to observe what's being oxidised and what's being reduced with the cell diagram. These are the chemical reactions that give rise to the cell's potential.

Cell Diagram 
Cell Diagram 

Cell Reactions 

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Oxidation reaction occurs: Ag → Ag2+ + 2e

Reduction reaction occurs: Cu2+ + 2e → Cu

The net cell reaction: Ag + Cu2+ ⇔ Ag2+ + Cu

Cell is represented as: Ag|Ag2+ || Cu2+|Cu


Difference between Cathode and Anode

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Cathode Anode
Because electrons are consumed, this is denoted by a positive sign. Since electrons are liberated here, it is denoted by a negative sign.
An electrochemical cell's cathode undergoes a reduction reaction. Here, an oxidation reaction takes place.
The cathode attracts electrons. Electrons are ejected from the anode.

Key to Determine a Cell's Potential

An electrochemical cell is depicted in the figure below. The gold-colored voltmeter at the very top measures the cell voltage, or the quantity of energy produced by the electrodes. The voltage of the electrochemical cell is determined by this voltmeter measurement. Ecell, or the potential difference between the half cells, is another name for this. Volts are the units of measurement for the amount of energy contained in each electrical charge; 1V=1J/C: V=voltage, J=joules, and C=coulomb. The electrons are essentially propelled by the voltage. When there is a high voltage, there is a lot of electron mobility. In Joules per Coulomb, the voltmeter measures the transfer of electrons from the anode to the cathode.

Key to Determine a Cell's Potential
Key to Determine a Cell's Potential

Standard Cell Potential

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The difference between the two electrodes, which creates the cell's voltage, is called the standard cell potential (E0cell). The following equation is used to find the difference between the two half cells:

E0Cell = E0Red, Cathode = E0Red, Anode

  • The standard cell potential is E0Cell.
  • The standard reduction potential for the reduction half-reaction occurring at the cathode is E0Red, Cathode.
  • The usual reduction potential for the oxidation half-reaction occurring at the anode is E0Red, Anode.

Things to Remember

  • The difference in potential between two half cells in an electrochemical cell is called the cell potential, Ecell.
  • The cell potential (Ecell) is measured in voltage (V), allowing us to assign a numerical value to it.
  • The cell diagram shows how the electrochemical cell's overall reaction works. It's easier to observe what's being oxidised and what's being reduced with the cell diagram. 
  • The cathode attracts electrons.
  • Electrons are ejected from the anode.
  • In Joules per Coulomb, the voltmeter measures the transfer of electrons from the anode to the cathode

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Sample Questions

Ques. Write out the oxidation and reduction half-reactions for this redox reaction. Create a cell diagram that corresponds to your equations. (2 marks)
Sn (s) + Pb2+ (aq) →  Sn2+ (aq) + Pb (s)

Ans. Oxidation reaction : Sn(s) → Sn2+(aq) + 2e-(aq)

Reduction reaction: Pb2+(aq) + 2e-(aq) → Pb(s)

Cell diagram: Sn(s) | Sn2+(aq) || Pb2+(aq) | Pb(s)

Ques. Which half cell has a higher potential to be decreased if Cu2+ ions in solution around a Cu metal electrode is the cathode and K+ ions in solution around a K metal electrode is the anode? (1 mark)

Ans. The half cell holding the Cu electrode in Cu2+ solution is the cathode, and it is in this half cell that reduction occurs. As a result, this half cell has a greater chance of being reduced.

Ques. What equipment is used to assess cell potential and how is it measured? (1 mark)

Ans. Volts (=J/C) are used to measure cell potential. A voltmeter can be used to determine this.

Ques. Describe Redox Reduction. (2 marks)

Ans. When one material is oxidised and another is reduced, this is known as a redox reaction. When a substance is oxidised, it loses one or more electrons and becomes positively charged. During reduction, on the other hand, the substance loses electrons and becomes negatively charged. 

Ques. What is Voltemeter Measurement? (1 mark)

Ans. The voltage of the electrochemical cell is determined by this Voltmeter Measurement.

Ques. What is the equation to find the difference between two half cells? (2 marks)

Ans. The following equation is used to find the difference between the two half cells:

E0Cell = E0Red, Cathode = E0Red, Anode

Ques. What is Ecell? (2 marks)

Ans. The difference in potential between two half cells in an electrochemical cell is called the cell potential, Ecell. It is measured in voltage (V), allowing us to assign a numerical value to it.

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CBSE CLASS XII Related Questions

1.

Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

      2.
      Write the Nernst equation and emf of the following cells at 298 K : 
      (i) Mg(s) | Mg2+ (0.001M) || Cu2+(0.0001 M) | Cu(s) 
      (ii) Fe(s) | Fe2+ (0.001M) || H+ (1M)|H2(g)(1bar) | Pt(s) 
      (iii) Sn(s) | Sn2+(0.050 M) || H+ (0.020 M) | H2(g) (1 bar) | Pt(s) 
      (iv) Pt(s) | Br2(l) | Br-  (0.010 M) || H+ (0.030 M) | H2(g) (1 bar) | Pt(s).

          3.

          Write equations of the following reactions: 
          (i)Friedel-Crafts reaction–alkylation of anisole.
          (ii)Nitration of anisole.

          (iii)Bromination of anisole in ethanoic acid medium.
          (iv)Friedel-Craft’s acetylation of anisole.

           

              4.
              Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

                  5.

                  Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. 
                  \((i) Methanal \)
                  \((ii) 2-Methylpentanal \)
                  \((iii) Benzaldehyde \)
                  \((iv) Benzophenone \)
                  \((v) Cyclohexanone \)
                  \((vi) 1-Phenylpropanone \)
                  \((vii) Phenylacetaldehyde \)
                  \((viii) Butan-1-ol \)
                  \((ix) 2, 2-Dimethylbutanal\)

                      6.
                      Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show: 
                       (i) Which of the electrode is negatively charged? 
                       (ii) The carriers of the current in the cell. 
                       (iii) Individual reaction at each electrode.

                          CBSE CLASS XII Previous Year Papers

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