Question

Balance the following redox reaction if it occurs in acidic solution:
Zn^2+ (aq) + NH4^+ (aq) --> Zn (s) + NO3^-

Answer 1

Can you show me your work ???

Answer 2

Hint : first break the equation in two parts

Answer 3

I dont know how to answer the question.

Answer 4

Zn^2+ 2electrons ----> Zn (1)
NH4^+ ----> NO3^- + 8 electrons (2)
first balance (1)

Answer 5

it is already balance , so balance the 2nd one

Answer 6

why is it NO3^- + 8 electron ? is'nt it suppose to be NO3^- + e

Answer 7

Hi. Are you still not clear about the electron part?

Answer 8

yes. Could you please help me with it ?

Answer 9

Sure. When solving these balancing redox reaction questions, you need to follow a few steps. They should work for all redox reactions (works for all questions I have encountered). First, you split the redox reaction into the reduction and the oxidation reactions, which you have already done:
\[Zn^{2+}\rightarrow Zn\]
\[NH_4 ^{+}\rightarrow NO_3 ^{-}\]

Answer 10

You shouldn't worry about electrons now, those come later. You then need to balance out the number of oxygen atoms on each side. In the reduction equation, there is no oxygen, so we can leave that alone for now. Let's look at the oxidation reaction. There are 3 oxygen atoms in NO3-, yet there is none on the ammonium side. So we add 3H2O on the side the lacks oxygen:

\[3H_2O+NH_4^{+}\rightarrow NO_3^{-}\]

Answer 11

See how now the number of oxygen atoms are balanced? Now we need to balance the number of hydrogen atoms. Again, the reduction equation doesn't have any hydrogen, so we will leave it alone. Looking at the oxidation reaction, there are 10 hydrogen atoms on the left side and there are none on the right. So we add 10 hydrogen ions on the right:
\[3H_2O+NH_4^{+}\rightarrow NO_3^{-}+10H^{+}\]

Answer 12

Now finally, it's time to balance the number of electrons on each side. Now this time, we don't leave the reduction equation alone, 'cause there is an imbalance of electrons on each side. To balance it out, we add electrons. So for the reduction:
\[2e^{-}+Zn^{2+}\rightarrow Zn\]
The net charge for each side is now 0. Now for the oxidation reaction. Looking at the reactants of the oxidation reaction from left to right: H2O has 0 charge. NH4 has a +1 charge. So this means +1 overall charge for left side. Then you have NO3, which has a -1 charge. And 10 hydrogen ions which have +10 charge. Net charge for right side is +9. We need to balance it out using the least amount of electrons. This means we need to add 8 electrons to the right side, so that both sides have an overall charge of +1:

\[3H_2O+NH_4^{+}→NO_3^{-}+10H^{+}+8e^{-}\]

Answer 13

That's why it's 8 electrons on the right.

Answer 14

But that doesn't answer the question, as you need to "recombine" these two equations together back to a full redox reaction. Do you know how to do this?

Answer 15

so, the full redox reaction would be : 4Zn^2+ (aq) + 3H2O (l) + NH4+ --> 4Zn (s) + 10H^+ (g)
Am I right ?

Answer 16

I think you forgot to put the NO3 on the right

Answer 17

Other than that, it looks fine to me.

Answer 18

ouh, yeah...thanks for helping.

Answer 19

No probs :)