Daniell Cell - Explanation, Diagram, Functions with FAQs

What is a Daniell Cell?

The Daniell cell is a type of electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction. It was developed by the English chemist John Frederic Daniell and is one of the most important examples used to understand galvanic cells. A Daniell cell consists of a zinc electrode dipped in zinc sulfate (ZnSO₄) solution and a copper electrode dipped in copper sulfate (CuSO₄) solution. The two half-cells are connected by a salt bridge, which helps maintain electrical neutrality by allowing the movement of ions between the solutions.

In this cell, zinc acts as the anode and undergoes oxidation:

$\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$

The electrons released travel through the external circuit to the copper electrode, where copper ions undergo reduction:

$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$

The overall cell reaction is:

$\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$

Explanation Of Daniell Cell

A conventional galvanic cell is meant to generate an electric current by using the spontaneous redox reaction between zinc and cupric ions.

A copper vessel makes up this cell. In this case, a saturated CuSO₄ solution is used as a depolariser and diluent. Fill with H₂SO₄, which works as an electrolyte. Zn₂SO₄ is used to submerge a zinc rod that has been amalgamated. CuSO₄ crystals are kept in touch with CuSO₄ solution by a transparent layer just below the upper surface of copper vessels, ensuring that the solution is always saturated.

Working Of the Daniell Cell

Daniel's cell working is discussed below.

$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

In a Daniell cell, electrons flow through an external circuit from the zinc electrode to copper electrode to copper electrode, while metal ions move from one half cell to the other via the salt bridge.

Through an external circuit, current flows from the copper electrode to the zinc electrode, which is the cathode, to the anode.

A voltaic cell can be reversible or irreversible, whereas a Daniell cell is reversible.

Cell Reaction

Chemical Reaction of Daniell Cells

$\mathrm{Zn}+\mathrm{Cu}^{2+} \rightleftharpoons \mathrm{Zn}^{2+}+\mathrm{Cu}$

Daniel's cell representation

$\mathrm{Zn} / \mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+} / \mathrm{Cu}$

The above reaction can be divided into two parts:

Half-cell anode reaction

$\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$

Zinc metal is oxidised once two electrons are liberated.

Reduction half-cell reaction
$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$

Copper metal is formed when copper ions are reduced to copper metal

Also, students can refer,

• NCERT solutions for Class 12 Chemistry Chapter 3 Electrochemistry
• NCERT Exemplar Class 12 Chemistry Solutions Chapter 3 Electrochemistry
• NCERT notes Class 12 Chemistry Chapter 3 Electrochemistry

Daniell Cell Construction

It's made up of a copper container filled with a concentrated copper sulphate solution. One porous cylindrical pot filled with diluted sulphuric acid is immersed in the concentrated copper sulphate solution inside the container. In a porous pot, one amalgam zinc rod is immersed in dilute sulphuric acid. Sulphuric acid in its diluted state has positive hydrogen ions and negative sulphate ions, which is a feature of dilute electrolytes.

When sulphate ions come into contact with zinc rods, they release electrons and generate zinc sulphate as a result of the oxidation reaction. As a result, the zinc rod acquires a negative charge and takes on the role of a cathode.

Positive hydrogen ions can pass through the porous wall of the pot and into the copper sulphate solution, where they combine with copper sulphate electrolyte sulphate ions to generate sulphuric acid. The positive copper ions in the copper sulphate electrolyte come into contact with the copper container's inner wall, where they are reduced and produce copper atoms, which are then deposited on the wall.

Salt Bridge

Salt bridge helps to maintain the electrical neutrality in two compartments by allowing movement of anions towards the anodic compartment and cations towards the cathodic compartment. A salt bridge is usually made by filling a U-tube or a tube with a gel or agar-agar containing an inert electrolyte, typically a salt like potassium nitrate ($\mathrm{KNO}_3$) or sodium chloride (NaCl). It's a gelatin-coated glass tube containing potassium chloride or ammonium nitrate.

Function of a Salt Bridge

Between the two half-cells, a salt bridge serves as an electrical contact.

• It restricts the mechanical movement of the solution but allows ions to migrate freely, allowing an electric current to pass through the electrolyte solution. It stops charges from building up.
• The charge balance in the two half-cells is maintained with the help of a salt bridge.
• The liquid connection potential is reduced or eliminated by using a salt bridge.

Also read -

Some Solved Examples

Question 1: Which of the following is true about chemical cells?

1) Convert chemical energy to electrical energy

2) Most batteries are chemical cells

3) (correct) i and ii

4) none

Solution:

Chemical Cells - The cells in which electrical energy is produced from the energy change accompanying chemical reactions or a physical process are known as chemical cells. Chemical cells convert chemical energy to electrical energy

Hence, the answer is option (3).

Question 2: Galvanisation is applying a coating of :

1) Pb

2) Cr

3) Cu

4) (correct) Zn

Solution:

As we learned

Exceptions of transition elements -

Zn, Cd, and Hg have the following configuration

$(n-1) d^{10} n s^2$ But they don't have empty 'd' orbitals that's why they are called 'd' block elements but not transition elements.

Zinc is used to prevent rusting.

Therefore, galvanisation is the application of a coating of Zn.

Hence, the answer is option (4).

Question 3: In a Daniell cell, which of the following occurs at the anode?

A) Reduction of $\mathrm{Cu}^{2+}$ to Cu
B) (correct) Oxidation of Zn to $\mathrm{Zn}^{2+}$
C) Oxidation of Cu to $\mathrm{Cu}^{2+}$
D) Reduction of $\mathrm{Zn}^{2+}$ to Zn

Solution:

At the anode, oxidation occurs. In the Daniell cell, zinc $(\mathrm{Zn})$ is oxidized to $\mathrm{Zn}^{2+}$ ions. The reaction is:

$\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(a q)+2 e^{-}$
So, zinc metal is oxidised, losing electrons.

Hence, the correct answer is option (B)

Question 4: A Daniell cell is represented as:

$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.01 \mathrm{M})\right|\left|\mathrm{Cu}^{2+}(1.0 \mathrm{M})\right| \mathrm{Cu}$
Given that:

$\begin{aligned}
& E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \\
& E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}
\end{aligned}$
The emf of the cell at 298 K is closest to:
(A) 1.04 V
(B) 1.10 V
(C) 1.16 V
(D) 1.22 V

Solution:

$E_{\text {cell }}^{\circ}=0.34-(-0.76)=1.10 \mathrm{~V}$
The cell reaction is:

$\begin{gathered}
\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu} \\
Q=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\frac{0.01}{1.0}=10^{-2}
\end{gathered}$
Using the Nernst equation:

$\begin{gathered}
E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log Q \\
=1.10-\frac{0.0591}{2} \log \left(10^{-2}\right) \\
=1.10-\frac{0.0591}{2}(-2) \\
=1.10+0.0591 \\
=1.1591 \mathrm{~V} \\
E_{\text {cell }} \approx 1.16 \mathrm{~V}
\end{gathered}$

Hence, the correct answer is option (C)

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