For the cell: Ni (s)| Ni^2+(0.1M)||Cu^2+(0.1M)|Cu(s) the cell potenti…

The correct answer is Option (3) → Cu²⁺ ion concentration is increased

The cell reaction and its Nernst equation (n = 2):

    \( \displaystyle \ce{Ni(s) + Cu^2+(aq) -> Ni^2+(aq) + Cu(s)} \)

    \( \displaystyle E_{cell} = E^\circ_{cell} - \frac{0.059}{2}\log\frac{[\ce{Ni^2+}]}{[\ce{Cu^2+}]} \)

Raising \( \ce{[Cu^{2+}]} \) (a reactant) makes the log term smaller, so E_cell increases.

Why the other options are incorrect

• Option (1) Ni²⁺ ion concentration is increased — Incorrect. \( \ce{Ni^{2+}} \) is a product; more \( \ce{Ni^{2+}} \) makes the log term larger and lowers E_cell.

• Option (2) Cu²⁺ ion concentration is decreased — Incorrect. Less reactant also lowers E_cell.

• Option (4) Temperature of the cell is decreased — Incorrect. Here \( \ce{[Ni^{2+}]} \) = \( \ce{[Cu^{2+}]} \) = 0.1 M, so the log term is zero and E_cell = E°_cell; lowering the temperature does not raise it through the Nernst term.

NCERT (2026-27 edition):
Nernst Equation – “… In Daniell cell, the electrode potential for any given concentration of Cu²⁺ and Zn²⁺ ions, we write
For Cathode: E_(Cu²⁺/Cu) = E°_(Cu²⁺/Cu) – (RT/2F) ln (1/[Cu²⁺(aq)])    (2.9)
For Anode: E_(Zn²⁺/Zn) = E°_(Zn²⁺/Zn) – (RT/2F) ln (1/[Zn²⁺(aq)])    (2.10)
The cell potential, E_(cell) = E_(Cu²⁺/Cu) – E_(Zn²⁺/Zn)
= E°_(Cu²⁺/Cu) – (RT/2F) ln (1/[Cu²⁺(aq)]) – E°_(Zn²⁺/Zn) + (RT/2F) ln (1/[Zn²⁺(aq)])
= E°_(Cu²⁺/Cu) – E°_(Zn²⁺/Zn) – (RT/2F) [ln (1/[Cu²⁺(aq)]) – ln (1/[Zn²⁺(aq)])]
E_(cell) = E°_(cell) – (RT/2F) ln ([Zn²⁺]/[Cu²⁺])    (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.”