How many grams of fluorine are needed to generate 7.65 moles Carbon Tetrafluoride
1 answer:
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<u>Answer:</u> The predicted cell potential of the cell is +0.0587 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u>
<u>Reduction half reaction (cathode):</u>
In this case, the cathode and anode both are same. So, will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
= 0.05 M
= 4.808 M
Putting values in above equation, we get:
Hence, the predicted cell potential of the cell is +0.0587 V