00:01
Continuing on with electrochemistry here, where we're essentially just looking at the movements of electrons throughout our system.
00:05
The first thing we're doing is calculating an e -cell value.
00:08
Where we have e -cell is the e -cathode reduction, subtract the e -anode oxidation.
00:16
E -cell is equal to 0 .446 volts, subtract 0 .242 volts.
00:24
E -cell is equal to 0 .204 volts.
00:30
And so the relationship between the cell potential and the free energy change.
00:36
Delta g equals negative n -f e -cell, where n is 2 moles, f is 96385.
00:46
E -cell is 0 .204.
00:48
So we can solve for delta g by plugging in all our numbers, why delta g is negative 3 .93 times 10 to the 4 joules.
00:59
Following this, we can take a look at the general formula of the announced equation, that is e -cell, it's equal to e -0, subtract 0 .0 .0591 divided by n, multiplied by log q, where q is cro 4, 2 minus, divided by h .e plus.
01:28
We have our products over our reactants...