Given the equilibrium constant: KC of the reaction: Cu(s) + Ag+ (aq) - Cu2+(aq) +2 Ag(s) is 10 × 1015, calculate the E0cell of this reaction at 298 K [2.303 (RT/F) textat 298 K =0.059 V]

Question

Given the equilibrium constant: KC of the reaction: Cu(s) + Ag+ (aq) -> Cu2+(aq) +2 Ag(s) is 10 × 1015, calculate the E0cell of this reaction at 298 K [2.303 (RT/F) textat 298 K =0.059 V] (A) 0.04736 V (B) 0.4736 V (C) 0.4736 mV (D) 0.04736 mV

Tardigrade Admin · Accepted Answer

E textcell =E° textcell - (0.059/n) log Q At equilibrium E° textcell = (0.059/2) log1016 = 0.059 ×8 = 0.472 V

Q. Given the equilibrium constant :
$K C$ of the reaction :
$Cu (s) + Ag X^{+} (aq) Cu X^{2 +} (aq) + 2 Ag (s)$ is $10 \times 1 0^{15}$ , calculate the $E_{ce ll}^{0}$ of this reaction at $298 K$ $[2.303 \frac{RT}{F} at 298 K = 0.059 V]$

0.04736 V

0.4736 V

0.4736 mV

0.04736 mV

$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} lo g Q$
At equilibrium
$E_{cell}^{\circ} = \frac{0.059}{2} lo g 1 0^{16}$
$= 0.059 \times 8$
$= 0.472 V$

Q. Given the equilibrium constant : KC of the reaction : Cu(s)+AgX+(aq)​CuX2+(aq)+2Ag(s) is 10×1015, calculate the Ecell0​ of this reaction at 298K [2.303FRT​at298K=0.059V]