Assume a cell with the following reaction $$\ce{Cu(s) + 2Ag^{+}(1 \times 10^{-3} M) -> Cu^{2+}(0.250 M) + 2Ag(s)}$$ $$E_{\text{cell}}^{\Theta}=2.97 \text{ V}$$ $E_{\text{cell}}$ for the above reaction is ______ V. (Nearest integer) [Given: $\log 2.5=0.3979$, $T=298 \text{ K}$]

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Redox Reactions and Electrochemistry

Assume a cell with the following reaction

Cu (s) + 2 Ag X^{+} (1 \times 10 X^{- 3} M) Cu X^{2 +} (0.250 M) + 2 Ag (s)

E_{cell}^{Θ} = 2.97 V

E_{cell}

for the above reaction is ______ V. (Nearest integer)
[Given:

lo g 2.5 = 0.3979

T = 298 K

]

Answer: 3

Two electrons pass for each copper atom oxidised, so

n = 2

, and the reaction quotient is

Q = \frac{[ C u ^{2 +} ]}{[ A g ^{+} ] ^{2}} = \frac{0.250}{( 1 \times 1 0 ^{- 3} ) ^{2}} = 2.5 \times 1 0^{5}

lo g Q = lo g 2.5 + 5 = 5.3979

. The Nernst equation at 298 K gives

E_{cell} = E_{cell}^{Θ} - \frac{0.059}{2} lo g Q = 2.97 - 0.0295 \times 5.3979 = 2.97 - 0.159

The cell potential is

2.81

V, which to the nearest integer is

3

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer