00:01
Answer for part a, the pairs of redox half reactions to be combined are iron, iron plus two, plus two electrons, and tellarium, te, plus two electrons to tellurium.
00:33
First, we take a look at a number of electrons released in an oxidation reaction and consumed in a reduction reaction and multiply the coefficient as needed for these numbers to become equal.
00:50
Since in this instance, the number of electrons in both equations is already equal.
00:58
We don't need to change the coefficients.
01:00
Next, we add together both reactions here.
01:06
The reaction became fe plus te, tolarium plus 2, and iron plus 2 plus 2 plus telarium, te.
01:21
Next we add water ions if it is needed to balance the equation.
01:27
In this instance, we don't need to do that because the previous equation is already balanced, therefore the final solution is here we have final solution f .e.
01:41
Plus tellurium plus 2, f .e.
01:49
Plus 2 plus terrarium.
02:05
Solution for part b, the pairs of redox of reactions to be combined are aluminum plus aluminum to aluminum to aluminum plus 3.
02:22
Plus three electrons and i -04 negative plus two electrons to i -0 -3 negative.
02:38
First we take a look at a number of electrons released in an oxidation reaction and consumed in a reduction reaction and multiply the coefficients as needed for these numbers to become equal.
02:51
In this instance, we multiply the coefficient in oxidation reaction by 2, this reaction by 2, and in reduction reaction by 3...