A Daniell Cell is an electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction. It is also referred as a galvanic or voltaic cell. It consists of two half-cells that are connected by a salt bridge to maintain electrical neutrality through the flow of ions. The first half-cell consists of a Zn rod dipped in a ZnSO4 solution, while the other half-cell consists of a Cu rod immersed in a CuSO4 solution. The two rods are connected by a wire to complete the circuit through which the electron flows, and a voltmeter may be connected in the circuit to measure the electrode potential. If the circuit in a Daniell cell represented by the cell notation, Zn | Zn2+ || Cu2+ | Cu, is closed, then it is observed that as time passes the concentration of Zn2+ keeps on increasing, while the concentration of Cu2+ keeps on decreasing. At the same time, the voltage of the cell as read on the voltmeter keeps on decreasing. After some time, the voltmeter shows zero reading.
Assume that at any instance of time, the reaction in the above mentioned Daniell cell has not attained equilibrium, then which one of the following statements is incorrect?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The electrode potential of the cell (Ecell) is always equal to its standard electrode potential.
Before equilibrium is reached, the cell potential at any instant is given by the Nernst equation (at 298 K, n = 2):
\( \displaystyle E_{cell}=E^{\circ}_{cell}-\frac{0.059}{2}\log\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
Ecell equals E°cell only when the log term is zero, i.e. when \( \ce{[Zn^{2+}]} \) = \( \ce{[Cu^{2+}]} \) (for example, both 1 M). As the cell works, \( \ce{[Zn^{2+}]} \) rises and \( \ce{[Cu^{2+}]} \) falls, so Ecell keeps changing.
Option (1) The electrode potential of the cell can be calculated by employing Nernst equation for the respective concentrations of Cu2+ and Zn2+ at that instance of time. — A correct statement about the cell before equilibrium. The Nernst equation gives Ecell for any concentrations.
Option (2) The electrode potential of the cell (Ecell) increases with the increase in the concentration of Cu2+ ions. — A correct statement about the cell before equilibrium. A larger \( \ce{[Cu^{2+}]} \) makes the ratio smaller, so less is subtracted from E°cell.
Option (3) The electrode potential of the cell (Ecell) increases with the decrease in the concentration of Zn2+ ions. — A correct statement about the cell before equilibrium. A smaller \( \ce{[Zn^{2+}]} \) also makes the ratio smaller, which raises Ecell.
Option (4) The electrode potential of the cell (Ecell) is always equal to its standard electrode potential. — The incorrect statement. Ecell equals E°cell only when the two ion concentrations are equal; as the cell works they change, so Ecell changes.
NCERT (2026-27 edition):
Nernst Equation – “…
\( \displaystyle E_{(cell)}=E^{\circ}_{(cell)}-\frac{RT}{2F}\ln\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \qquad (2.11) \)
It can be seen that E(cell) depends on the concentration of both Cu2+ and Zn2+ ions. It increases with increase in the concentration of Cu2+ ions and decrease in the concentration of Zn2+ ions.”