E naught cell for the following galvanic cell is +0.254V
Hg2^2(aq)+2I^-(aq) ==> 2 Hg(l) + I2(s)
What is delta G naught for this reaction?
Delat G naught = -nFEnaught
n = the number of moles of electrons transferred between oxidizing and reducing agents as determine by the balanced equation.
-.254x96,500 = -24511kJ (I wrote down the value that you want me to use for Faraday, but I can't find it right now)
To calculate ΔG° for the reaction, we need to know the number of moles of electrons transferred (n) and Faraday's constant (F).
The balanced equation shows that 2 moles of electrons are transferred in the reaction:
Hg2^2+(aq) + 2I^-(aq) → 2Hg(l) + I2(s)
So, in this case, n = 2.
Faraday's constant (F) is approximately 96,500 C/mol.
Now we can calculate ΔG°:
ΔG° = -nFE°
Substituting the values:
ΔG° = -(2)(96,500 C/mol)(0.254 V)
ΔG° ≈ -49,110 J/mol or -49.11 kJ/mol (since 1 kJ = 1000 J)
Therefore, ΔG° for this reaction is approximately -49.11 kJ/mol.