Given the balanced ionic equation:
zn(s) + cu2+(aq) → zn2+(aq) + cu(s)
which equation represents the oxidation half reaction?
(1) zn(s) + 2e– → zn2+(aq)
(2) zn(s) → zn2+(aq) + 2e–
(3) cu2+(aq) → cu(s) + 2e–
(4) cu2+(aq) + 2e– → cu(s)

Option (2) is the correct answer.

Explanation:

Oxidation is the process in which there is loss of electrons or increase in oxidation number of the element.

Whereas reduction is the process in which there is gain of electrons or decrease in oxidation number of the element.

For example,

Oxidation-half reaction:

Reduction-half reaction:

Thus, we can conclude that the given equation Zn(s) → Zn2+(aq) + 2e– represents the oxidation half reaction.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

i believe that it is a exoplanet