Daniell Cell
Electrochemistry is the study of the production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations. The cell below represents Daniell cell which converts the chemical energy liberated during the redox reaction to electrical energy.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
The reaction is a combination of two half-cell reactions, one called the oxidation half-cell and the other called the reduction half-cell. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected through a metallic wire and switch externally. The electrolyte of the two half cells is connected internally through a salt bridge. At each electrode-electrolyte interface there is a tendency for metal ions from the solution to deposit on the metal electrode, trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode, trying to make it negatively charged. A potential difference develops between the electrode and the electrolyte, which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity, then the electrode potential is known as the standard electrode potential.
For a general half-cell; Mn+(aq) + ne− → M(s), the Nernst equation is written as
\( E_{(M^{n+}/M)}=E^{o}_{(M^{n+}/M)}-\frac{RT}{nF}\ln\frac{1}{[M^{n+}]} \)
where \( E^{o}_{(M^{n+}/M)} \) is the standard reduction potential of the half-cell, R is gas constant (8.314 J K−1 mol−1), F is Faraday constant (96487 C mol−1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn+.
Calculate the emf of the Daniell cell when the concentration of [Cu2+] and [Zn2+] is unity.
[Given: The Standard electrode potential for the Daniell cell is 1.1V]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.1 V
The emf of a cell under any conditions is given by the Nernst equation. For the Daniell cell (n = 2) at 298 K:
\( \displaystyle E_{cell}=E^{o}_{cell}-\frac{0.059}{2}\log\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
When the concentrations of all ions are unity, the ratio inside the log is 1, and log 1 = 0. The cell is then under standard conditions, and its emf equals the standard value:
\( \displaystyle E_{cell}=1.1-\frac{0.059}{2}\log\frac{1}{1} \)
\( \displaystyle E_{cell}=1.1-0=1.1\ \mathrm{V} \)
So the emf of the Daniell cell is 1.1 V.
NCERT (2026-27 edition):
Electrochemical Cells
