00:02
For a, we have a degree sub cell equals 0 .54v minus 1 .09v.
00:12
This equals negative 0 .545v.
00:19
And thus, it's favored.
00:25
B, actually, for a, we'll also add i2s plus 2br negative aq, 2i negative aq plus br2.
00:52
And this as well.
00:56
For b, we have cut positive aq plus 2fe2 positive aq, cus plus 2fe3 positive aq, and negative 0 .43v.
01:27
And we can see it's favored.
01:30
For c, we have 6fe2 positive aq plus cr2 of 7, 2 negative aq plus 14h positive aq.
01:54
And we have 6fe3 positive aq plus 2cr3 positive aq plus 7h2o...