Explanation:
The given chemical reaction is:


The relation between Eo cell and Keq is shown below:

The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,

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:

Answer:
Keq=6.13x10^33
2BF₃ + 3Li₂SO₃ ----> B₂(SO₃)₃ + <u>6LiF
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Where is the link can’t do any work without it
First blank -Same
Second blank-Neutral