Electrochemical cell | مهار فن ابزار

An electrochemical cell is a device that produces electrical energy through chemical reactions inside it . In any electrochemical process , electrons flow from one chemical to another through an oxidation-reduction reaction. An oxidation-reduction (redox) reaction occurs when electrons are transferred from an oxidized substance to a reducing substance. In this type of process, “reductant” is called a substance that loses electrons and is oxidized. “Oxidant” is also referred to as particles that are reduced during the process by accepting electrons.

Table of contents of this article

half-reactions in the electrochemical cell

Types of electrochemical cells

Galvanic cell

Electrochemical cell diagram

Among the common examples of electrochemical cells, we can mention the 1.5 volt battery , which is used in many electronic devices. Such cells, which have the ability to generate electric current with the help of chemical reactions, are called ” galvanic cells ” or voltaic cells. On the contrary, the electrochemical cell in which chemical reactions are carried out with the help of applying an electric current, are included in the category of electrolytic cells . In the picture below, the general shape of the components of an electrochemical cell is shown.

Note that when displaying an electrochemical cell, the cathode is displayed on the right and the anode is displayed on the left.

half-reactions in the electrochemical cell

Oxidation and reduction reactions can be represented as two “half-reactions”. One of these half-reactions represents the reduction process and the other represents the oxidation process. For example, the reaction of zinc with bromine is shown as follows:

$Z n (s) + B r_{2} (a q) \to Z n^{2 +} (a q) + 2 B r^{-} (a q)$

This reaction can be shown as two half-reactions:

Reduction half-reaction: $B r_{2} (a q) + 2 e^{-} \to 2 B r^{-} (a q)$

Oxidation half-reaction: $Z n (s) \to Z n^{2 +} (a q) + 2 e^{-}$

Each of these half-reactions is written to better understand the process. In this reaction, zinc, reducing and $B r_{2}$ It is oxidizing. By adding the above two equations, we get the overall reaction. In many chemical reactions, we assume that the reactants are physically in contact with each other. For example, in acid and base reactions , both substances are dispersed in the same phase. But in the case of oxidation-reduction (redox) reactions, half-reactions can be physically separated from each other. Of course, to separate these half-reactions, we need a complete circuit.

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To complete the circuit and connect the two half-reactions, we use an external wire. As the reaction progresses, electrons flow from the reducing agent to the oxidizing agent with the help of an electric wire. As a result of this process, electrical energy is created with the help of which work can be done. A device that uses a spontaneous oxidation-reduction reaction to produce electrical energy is called an electrochemical cell. Of course, there is another type of electrochemical cell that uses electrical energy to perform the non-spontaneous oxidation-reduction reaction.

Types of electrochemical cells

There are 2 different types of electrochemical cells. These cells are called “galvanic cells” and electrolytic cells. Galvani cells are named after the Italian physicist Luigi Galvani. During the autopsy of a frog, this scientist observed that when an electric shock was applied, the legs of this frog moved and this indicated the electrical nature of nerve messages .

Galvanic cell

A galvanic cell produces electrical energy by using the energy released during a spontaneous oxidation-reduction reaction. This type of electrochemical cell is also known as voltaic cells, named after its inventor Alessandro Volta.

Electrolytic cell

On the other hand, an electrolytic cell, with the consumption of electrical energy supplied from an external source, causes non-spontaneous reactions. $(Δ G > 0)$ Oxidation is reduced. Each of these two types of cells includes two electrodes. These electrodes form solid metals that are connected to an external circuit, and this circuit is responsible for connecting the two parts of the cell.

The oxidation half-reaction occurs at one electrode (anode) and the reduction half-reaction occurs at the other (cathode). By completing (closing) the circuit, electrons flow from the anode to the cathode. These electrons are also bound together by an ionic substance or solution. This substance or ionic solution causes the transfer of electrons from one part of the electrode to another and is called an electrolyte .

Galvanic cell

In the following, we will examine the principles of galvanic cells (voltaic) and its reactions. For this purpose, we consider the zinc-copper electrochemical cell (the reaction of zinc metal with copper ions), which ultimately leads to the production of copper and zinc ion deposits. Its balanced reaction is given below.

$Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$

To perform such a reaction, we can place a piece of a zinc ingot in an aqueous solution of copper (II) sulfate. As the reaction progresses, the zinc ingot dissolves and a mass of copper is formed. These changes happen spontaneously, but all the energy released is released as heat and cannot be used to do work.

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Such a reaction can be done in a galvanic cell. To prepare this cell, put a strip of copper inside a beaker containing a 1 M solution of ions $C u^{2 +}$ and put a strip of zinc metal inside the molar solution of the ions $Z n^{2 +}$ Put. Two metal strips play the role of electrodes and are connected to each other by a wire. Containers containing solutions are also connected to each other with the help of a salt bridge . In fact, this salt bridge is a U-shaped tube that is placed inside human beings.

Salt bridge ions are selected in such a way that they do not participate in electrochemical reactions and do not undergo oxidation or reduction and do not form complex deposits and compounds . Common cations and anions used in salt bridges are respectively sodium or potassium ions and $N O_{3}^{-}$ Or $S O_{4}^{2 -}$ . When the circuit is closed, a spontaneous reaction occurs. In the anode (zinc electrode), zinc metal to $Z n^{2 +}$ oxidizes and in the cathode (copper electrode), ions $C u^{2 +}$ They are reduced to copper metal.

As the reaction progresses, the zinc strip will dissolve and the concentration of ions $Z n^{2 +}$ in a solution containing $Z n^{2 +}$ increases At the same time, the mass of the copper strip increases and the concentration of ions $C u^{2 +}$ It decreases in the solution containing these ions . In this way, we repeated the same reaction as above in a single human by separating the oxidation and reduction half-reactions. The electrons released in the anode flow along the wire and produce an electric current. As a result, galvanic cells convert chemical energy into electrical energy that can be used to do work.

Cell components on copper (Click on the image to view it in a larger size.)

The electrolyte inside the salt bridge has two main properties, which we will discuss further.

It completes the circuit by carrying electrical charges and maintaining a neutral electrical state in both solutions, which occurs through the migration of ions across this salt bridge. As long as the salt bridge ions do not participate in redox reactions under the operating conditions of the cell, their chemical nature in the salt bridge does not matter. Without such binding, the overall positive charge in the solution $Z n^{2 +}$ , with the dissolution of zinc metal, it increased and the overall positive charge in the solution $C u^{2 +}$ was decreasing

Salt bridge through the flow of anions into the solution $Z n^{2 +}$ and cations to the solution $C u^{2 +}$ , it causes loads to be neutralized. In the absence of a salt bridge or similar junctions, the reaction would proceed rapidly due to the non-maintenance of the neutral electrical state.

A voltmeter can be used to measure the potential difference between two people. If we put a switch in the path of the wire, when the switch is cut (the path connecting the anode to the cathode is cut), the current will stop, and as a result, we will not have any chemical reaction. By connecting the key, the external circuit is completed and the flow from anode to cathode is possible.

cell potential

cell potential $(E_{c e l l})$ Expressed in volts, it is called the electric potential difference between two half-reactions, which is related to the energy required to move a charged particle in an electric field. In the cell we explained, the voltmeter shows a potential difference equal to 1/10 volts. Since the electrons of the oxidation half-reaction are released at the anode, the anode in the galvanic cell will have a negative charge and the cathode – which absorbs the electrons – will have a positive charge.

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Note that not all electrodes undergo chemical transformations during a redox reaction. The electrode can be made of a neutral and highly conductive metal such as platinum to prevent its reaction in the oxidation-reduction process. We have examined such a phenomenon in the following example.

Galvanic cell example

A chemist has prepared a galvanic cell consisting of two electrodes. One probe consists of a strip of tin immersed in an aqueous solution of sulfuric acid , and the other has a platinum electrode in a solution of nitric acid. Both solutions are connected to each other through a salt bridge and the electrodes are connected with the help of an electric wire. When the circuit is connected, gas bubbles appear in the platinum electrode. The following balanced equation represents the spontaneous oxidation-reduction reaction in this electrochemical cell.

$3 § n (s) + 2 N O^{- 3} (a q) + 8 H^{+} (a q) \to 3 § n^{2 +} (a q) + 2 N O (g) + 4 H_{2} O (l)$

Consider the following for this galvanic cell:

Write the half-reaction of each electrode.
Show which electrode is the cathode and which is the anode.
Indicate which electrode is positive and which is negative.

To solve this question, after determining the oxidation and reduction half-reactions in each electrode, we determine the sign of the cathode and anode with the help of the direction of the flow of electrons.

In the reduction half-cell, nitrate is converted to nitric oxide during the reduction reaction. Next, this nitric oxide reacts with oxygen in the air $N O_{2}$ which is characterized by red-brown color. In the oxidation half-reaction, tin metal is oxidized. The half-reactions corresponding to the described processes are given below:

Reduction half-reaction: $N O_{3}^{-} (a q) + 4 H^{+} (a q) + 3 e^{-} \to N O (g) + 2 H_{2} O (l)$

Oxidation half-reaction: $§ n (s) \to § n^{2 +} (a q) + 2 e^{-}$

Therefore, nitrate to $N O$ decreases and tin to $§ n^{2 +}$ It is oxidized. Since the reduction occurs at the platinum electrode, so our cathode in this cell is platinum. Similarly, due to the occurrence of oxidation in the tin electrode, the anode will be the tin electrode in this electrochemical cell.

Electrons flow from the tin electrode to the platinum electrode through an electrical wire and are transferred to the nitrate at the platinum electrode. The electrical circuit is completed with the help of a salt bridge. This bridge is the cause of the passage of cations to the cathode and anions to the anode. Since electrons flow from the tin electrode, this electrode will be our negative electrode. On the contrary, due to the flow of electrons towards the platinum electrode, we consider the platinum electrode as the positive electrode.

The second example is a galvanic cell

Consider a simple galvanic cell in which 2 cells are connected by a salt bridge. A beaker containing the solution $M n O_{4}^{-}$ in dilute sulfuric acid with a platinum electrode. In other humans, a solution of $S n^{2 +}$ We have dilute sulfuric acid. The electrode of this half cell is considered to be platinum. When we connect two electrodes with a wire, the following reaction takes place.

$2 M n O^{-} 4 (a q) + 5 S n^{2 +} (a q) + 16 H^{+} (a q) \to 2 M n^{2 +} (a q) + 5 S n^{4 +} (a q) + 8 H_{2} O (l)$

As in the previous example, write the half-reactions of each electrode and specify the cathode and anode along with their symbols.

The half-reactions will be as follows:

$\begin{aligned} M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} & \to M n^{2 +} (a q) + 4 H_{2} O (l) \\ S n^{2 +} (a q) & \to S n^{4 +} (a q) + 2 e^{-} \end{aligned}$

The platinum electrode in the permanganate solution is the cathode and the tin electrode is also the anode. According to the current direction, the cathode will be positive and the anode will be negative.

Electrochemical cell diagram

If you have read this article thoroughly, you will see that every time we describe an electrochemical cell, we have to give lengthy explanations to describe the half-cells, electrodes, and electrolyte. To avoid additional explanations, we can use a linear representation called “Cell Diagram” to represent the electrochemical cell. In this type of representation, we show the chemical nature of the electrodes and parts of the cell with the help of the chemical formula , and the anode is placed on the left side and the cathode is placed on the right side.

The boundary between the phases is displayed with the help of vertical lines (a vertical line). We also show the salt bridge through two vertical lines. Therefore, we can display the cell diagram for the electrochemical cell on copper as shown below:

Galvanic cells can have configurations other than the examples we have examined so far. For example, the voltage resulting from an oxidation-reduction reaction can be measured more accurately by placing two electrodes in an electrolyte test tube. This type of arrangement reduces the errors caused by the resistance of the phase boundary for the flow of loads, which is called “junction potential” or connection potential. An example of this galvanic cell is shown in the diagram below.

$P t (s) | H_{2} (g) | H C l (a q) | A g C l (s) A g (s)$

With a little attention, you will find that this electrochemical cell diagram does not have vertical lines, that is, we do not have a salt bridge in this cell, for this reason, we have not included two vertical lines in the diagram. The half-reactions and the overall reaction are as follows:

Reaction at the cathode: $A g C l (s) + e^{-} \to A g (s) + C l^{-} (a q)$

Reaction at the anode: $\frac{1}{2} H_{2 (g)} \to H_{(a q)}^{+} + e^{-}$

General reaction: $A g C l_{(s)} + \frac{1}{2} H_{2 (g)} \to A g_{(s)} + C l_{(a q)}^{-} + H_{(a q)}^{+}$

A single-section galvanic cell produces the same voltage as the two-section cell, but discharges it faster. The reason for this rapid discharge is the direct reaction of the reactants in the anode. For this reason, such a cell is not suitable for generating electricity.

An example of an electrochemical cell diagram

Draw the cell diagram for the galvanic cell of the previous example. Its balanced equation is as follows:

$3 § n (s) + 2 N O_{3}^{-} (a q) + 8 H^{+} (a q) \to 3 S n^{2 +} (a q) + 2 N O (g) + 4 H_{2} O (l)$

The anode of this cell is a strip of tin and its cathode is platinum. Starting from the left, we use the vertical line to show the phase boundary between the electrode and the tin-containing solution. Therefore, we will write the anode part as follows:

$§ n (s) ∣ § n^{2 +} (a q)$

In this type of display, we could say $H_{2} § O_{4} (a q)$ Use it in the cathode section, but since this compound does not participate in the overall reaction, there is no need to display it. The cathode part contains nitric acid solution along with the reaction product $(N O)$ And platinum participates in the reaction. We write this section as follows, where the vertical lines represent the boundaries of the phases. The

$H N O 3 (a q) ∣ N O (g) ∣ P t (s)$

By combining these two parts in the electrochemical cell, we get the following diagram:

$§ n (s) | § n^{2 +} (a q) | | H N O_{3} (a q) | N O (g) | P t_{(} s)$

Considering that the concentrations of the solution in the question are not given, we do not display them in the diagram. In the next example, we show the cell diagram along with the concentration of the solutions.

The second example of drawing a diagram of an electrochemical cell

Draw the cell diagram for the following reaction. Suppose that in this cell, the concentration of silver and magnesium ions is equal to 1 M.

$M g (s) + 2 A g^{+} (a q) \to M g^{2 +} (a q) + 2 A g (s)$

Finally, according to the previous example and the given concentrations, its diagram is drawn as follows:

$M g (s) | M g^{2 +} (a q 1 M) | | A g^{+} (a q 1 M) | A g (s)$

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half-reactions in the electrochemical cell

Types of electrochemical cells

Galvanic cell

Electrolytic cell

Galvanic cell

cell potential

Galvanic cell example

The second example is a galvanic cell

Electrochemical cell diagram

An example of an electrochemical cell diagram

The second example of drawing a diagram of an electrochemical cell

2 thoughts on “Electrochemical cell”

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