Redox Reactions

Redox Reactions and Electrode Potential

We come across many things in our daily life that are related to electrochemical reactions directly or indirectly. Flashlight and batteries involve electrochemical reactions and use the principle of an electrochemical cell.  An electrochemical/Daniell/ Voltaic cell is related to electrode potential. How? Can you name a few more modern-day gadgets where an electrochemical cell or Daniell cell forms the principle? Let us find out in this topic.

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Galvanic Cell/ Voltaic Cell

It is an electrochemical cell that uses redox reaction or redox process to produce electrical energy/electricity. Refer to the diagram below to understand a voltaic cell.

Electrode Potential

Galvanic/Voltaic Cell (Source Credit: Wikipedia)

A voltaic cell consists of two different compartments. The two separate parts are the half-cells of the voltaic cell where either the reduction process or the oxidation process will take place. The two half-cells are the left and right half-cell. According to the diagram shown above, the right half-cell consists of a copper metal rod/strip in a copper (II) sulfate solution.

The left half-cell consists of a zinc metal rod/strip in a zinc sulfate solution. The metal strips are electrodes. It acts as a conductor in the complete circuit arrangement and helps in the transfer of electrons to a non-metallic part present in the arrangement. The non-metallic part is the electrolyte solutions where the strips of metals acting as an electrode are present.

In the above diagram, the two electrodes are connected by a metal wire. A switch helps to open and close the circuit. The porous membrane present between the two parts (half-cells) helps to complete the circuit arrangement. The best way to describe the electrochemical process in the voltaic cell is by the example of a zinc-copper cell. So let us understand the redox reaction of zinc and copper.

Redox Process (Oxidation & Reduction Reaction) of Zinc & Copper

Considering the above diagram, the zinc rod/strip is oxidized to zinc (Zn2+) ions by direct electron transfer. Similarly, copper ions are reduced to the copper element by direct electron transfer. We know from above experiment, the connection of zinc and copper rods are external in order to complete the circuit and help in the transfer of electrons. The two half-reactions are as follows:

Oxidation Half-Reaction

Zn (s) → Zn2+ + 2e 

Reduction Half-Reaction

Cu2+ + 2e→ Cu(s) 

The element zinc loses electrons which the copper ions acquire to form metallic copper. Therefore, the overall redox reaction of zinc and copper is:

Cu2+ +   Zn (s)  →   Cu (s)   +   Zn2+

Observation of the Redox Process in Zinc and Copper Cell

Zinc electrode release zinc atoms which get oxidized to its ionic form. This is because zinc is higher in the activity series in comparison to copper. Therefore, zinc is oxidized more easily than copper.

Zn (s)   →   Zn2+   +   2e 

The electrode where oxidation takes place is called anode. In this case, zinc anode slowly depreciates due to loss of zinc metal while the cell operates. Thus, the concentration of zinc ion will gradually increase because of electron generation at the anode. Therefore, it is the negative electrode.

The electron moves from zinc anode through the external wire and continues towards the copper electrode. As a result, voltmeter registers a reading. Finally, the electrons will enter the copper electron and combine with copper ions present in the solution to form the metallic copper.

Cu2+ (aq)   +   2e–   →    Cu(s) 

The reduction takes place in the copper electrode. It is also called a cathode. Gradually the mass of cathode will increase because of the formation of copper metal. However, the half-cell concentration of the copper (II) ions will decrease. Thus, cathode acts as a positive electrode.

The movement of the ions through the membrane helps in maintaining the neutrality in the cell. In the diagram of voltaic cell mentioned above, the sulfate ions will travel from the copper half-cell to the zinc half-cell to maintain the increase and decrease taking place in the copper and zinc ions, respectively.

Daniel Cell

The diagram shown below represents a Daniel Cell. It is same as an electrochemical/voltaic cell.

Electrode Potential

Daniel Cell (Source Credit: Wikipedia)

The experiment is exactly the similar to the Voltaic cell where we take zinc rod in a beaker of zinc sulfate solution and copper rod in a beaker of copper sulfate solution for the indirect transfer of electrons. Oxidation and reduction will occur in two half-cells of zinc and copper. A redox couple will take place.

In a redox couple, the substance undergoing oxidation and reduction is present together. In a redox couple, a vertical line acts as a separation or an interface between the oxidized and reduced form. The interface can be solid or in solution form.

In a Daniell cell, the redox couple is Zn2+/Zn and Cu2+/Cu. We keep the two beaker containing the zinc sulfate and copper sulfate solution side by side and connect them with the help of a salt bridge. Salt bridge is nothing but a tube in a U form having a solution of ammonium nitrate or potassium chloride. Initially, the solution is solidified by boiling it with agar-agar and then cooling it until it forms a jelly-like substance.

The salt bridge acts as the electric contact between the solutions. At the same time, it helps to keep the solution separate. The copper and zinc rods are joined by a metallic wire, ammeter, and a switch. This complete set up is an example of Daniel Cell. No reaction takes place and there is no current flow when the switch is off. However, the moment the switch starts we can observe the following things

Observations of Daniel Cell

The electronic transfer occurs through the metallic wire connecting the rods. The direction of the arrow indicates the flow of current. The flow of electricity occurs due to the migration of ions from one solution of the beaker to the other by the help of salt bridge. However, the flow of current is not possible until there is a potential difference between the two electrodes (copper and zinc rods).

Electrode Potential

We know that the half-reactions occur at the electrodes. Thus, summarizing the experiment

  • Anode- The electrode where oxidation occurs
  • Cathode- The electrode where reduction occurs
  • Electrode Potential- Overall potential occurring in each electrode

Standard Electrode Potential

The potential related to each electrode refers to elected potential. If the species concentration participating in the electrode reaction is unity and the reaction occurs at 298 K, then the potential of the electrode refers to as Standard Electrode Potential (E0). The electrode reaction is at unity when any gas participating in the particular electrode reaction is approximately 1 atmospheric pressure.

Considering the case of the convention, the standard electrode potential (E0) of hydrogen gas is 0.00 volts. The electrode potential value of a particular electrode process indicates the measure of the relative tendency of any active species in the process to remain in its redox form.

If the value of standard electrode potential is negative, this signifies the redox couple is the stronger reducing agent in comparison to H+ /H2 couple. Similarly, a positive standard electrode potential signifies that the redox couple is a weaker reducing agent in comparison to H+ /H2 couple. There is a specific table of standard electrode potential or electrode potential at 298 K.

Calculation of Electrode Potential

The formula for calculation of electrode potential is,

E0cell = E0red − E0oxid

We can follow the above-given link for the standard electrode potential table for the determination of reactions that will take place and for the determination of standard cell potential for any number of combinations of two half-cells. This process will not even require an actual construction of the cell.

The half-cell having the higher reduction potential as per the table will undergo reduction process in the particular cell. Similarly, the half-cell having lower reduction potential will undergo oxidation process inside the particular cell. Following the specifications, the determination of overall cell potential is possible. It will always be a positive value otherwise the redox reaction will not be spontaneous. The negative cell potential indicates that the reaction will be spontaneous but in reverse direction.

Uses of Electrode Potential

  • It helps to study the processes such as crevice corrosion and pitting which further helps to control the reaction process.
  • Electrode Potential is useful for choosing materials and devices that help in the control of the reaction process.
  • Helps in the prediction of the corrosion related to the electrochemical and chemical reactions/ processes.

A Solved Question for You

Q: By using the value of standard electrode potential, determine whether a reaction is possible in the below-mentioned species.

  • Fe3+(aq) and I(aq)
  • Fe3+ (aq) and Br(aq)
  • Br2 (aq) and Fe2+ (aq)

Solution: For a reaction to be feasible, Eo cell will always be positive. We know that, ∆rGθ < 0 and ∆rGθ = –nFEo cell. Therefore, –nFEo cell < 0. In the given formula, n and F are positive values. Thus, –Eo cell < 0. Changing the sign on the equation we get, Eo cell > 0. Hence,  Eo cell will always be positive in the case of a feasible reaction.

  • Fe3+ (aq) and I(aq): Balance half reactions and the respective standard electrode potential of Fe3+ (aq) and I(aq) are

2Fe3+ +  2e →   2Fe+2 (Eo) = +0.77 V

2I →   I2 + 2e  (Eo) = -0.54 V

Adding the values we get: (Eo)= +0.23. Therefore, Eo cell > 0 and the reaction is possible.

  • Fe3+ (aq) and Br(aq): Possible half reactions and the respective standard electrode potential of Fe3+ (aq) and Br(aq)

2Fe3+ + 2e   →   2Fe+2 (Eo) = +0.77 V

2Br  →   Br2 + 2e (Eo) =  –1.09 V

Addition of the values results in (Eo)= -0.32 V. Therefore, Eo cell < 0 and the reaction is not possible.

  • Br2 (aq) and Fe2+ (aq): Complete balance half reactions and the respective standard electrode potential of Br2 (aq) and Fe2+ (aq)

Br2 + 2e    →    2Br (Eo) = + 1.09 V

2Fe+2   →    2Fe3+ +  2e (Eo) =  – 0.77 V

Adding the values Eo will be: (Eo) = + 0.32 V. Therefore, Eo cell > 0. Hence, the reaction is possible.

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