The emf of cell $\mathrm{T} 1\left|\underset{(0.001 \mathrm{M})}{\mathrm{T} 1^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}$ is 0.83 V at 298 K . It could be increased by :

(D) increasing concentration of $\mathrm{Cu}^{2+}$ ions

(A) increasing concentration of $\mathrm{Tl}^{+}$ions

(B) increasing concentration of both $\mathrm{Tl}^{+}$and $\mathrm{Cu}^{2+}$ ions

(C) decreasing concentration of both $\mathrm{Tl}^{+}$and $\mathrm{Cu}^{2+}$ ions

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Redox Reactions and Electrochemistry

The emf of cell

T 1 (0.001 M) T 1^{+} (0.01 M) Cu^{2 +} ∣ Cu

is 0.83 V at 298 K . It could be increased by :

Answer: (D)

Cell reaction:

2 T l (s) + C u^{2 +} (a q) \to 2 T l^{+} (a q) + C u (s)

n = 2

E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.0591}{2} lo g \frac{[ T l ^{+} ] ^{2}}{[ C u ^{2 +} ]}

E_{ce l l}

rises as the log term falls, which happens on increasing

[C u^{2 +}]

(denominator grows) or decreasing

[T l^{+}]

. Among the choices offered, increasing

[C u^{2 +}]

ions increases

E_{ce l l}

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option