Question

Determine the equilibrium constant, Keq, at 25°C for the reaction

Answer 1

Explanation:

The given chemical reaction is:

$2Br^- (aq) + I_2(s) Br_2(l) + 2I^- (aq)$

$E^ocell=oxidation potential of anode + reduction potential of cathode$

The relation between Eo cell and Keq is shown below:

$deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell$

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

$E^ocell= (-1.07+0.53)V\\=-0.54V$

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

$ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3$

Answer:

Keq=6.13x10^33