Question

Balance the equation using the half-reaction method:
Sn + Ag^+1 (arrow) Sn^2 + Ag

Answer 1

Break this down into two half reactions, one for loss of electrons (Ox) and the other for gain for electrons (Red). Then balance it by making sure that the number of electrons lost =number of electrons gained - adjust coefficients of the reactants and products.

Answer 2

Can someone please help me!!!!!

Answer 3

Allright i'll explain in detail what @Preetha said yesterday.
First,
You have 2 half reactions, 1 oxidation and 1 reduction.
Oxidation = loss of electrons
Reduction = gain of electrons
Because electrons are negatively charged, the oxidation state will decrease with a reduction and the oxidation state will increase with reduction. So when Sn loses 2 electrons \((2e^-)\) the oxidation state goes from \(Sn^0\) to \(Sn^{2+}\)

Do you understand this?

Answer 4

hhhmmmmm.....

Answer 5

so the this reaction is reduced

Answer 6

This reactions contains a reduction and an oxidation. You have to find out which element gets reduced and which one gets oxidized. To do this you need to find out which element loses electrons and which one gains electrons.

Answer 7

First, do i have to find the oxidation # for Sn? And then figure out if it is reduced?

Answer 8

This is the reaction: \(\Large\sf Sn^0+Ag^+\to Sn^{2+}+Ag^0\)
I added the zeros to make it more simple.
Let's start with tin (Sn)
On the left side of the reaction (before the arrow) the tin atom has an oxidation state of zero. After the arrow it has an oxidation state of 2+
Does it lose or gain electrons?

Answer 9

gain. so its reduced

Answer 10

Is Ag oxidized?

Answer 11

Sn actually Loses electrons. So its oxidized. Its a bit confusing.
When the oxidation state increases, it means it loses electrons.

It's more easier to explain it for silver
Before the reaction it has an oxidation state of +1 and after the reaction 0
This means it gains 1 electron. Because an electron is negatively loaded, it needs to gain 1 electron to get an oxidation state of 0.
So you would get a half reaction: reduction = \(\Large\sf Ag^+ +e^- \to Ag^0\)

Answer 12

+1 -1 = 0, do you understand that?

Answer 13

Ok .. I get it!

Answer 14

Okey! now for the tin.
What would be the half reaction for the oxidation?

Answer 15

umm...

Answer 16

it gains 2 electrons so its reduced

Answer 17

when you see it gains 2+, it means it loses 2 electrons.

Answer 18

right! its oxideized!

Answer 19

exactly.
So whats the half reaction for the oxidation?

Answer 20

Sn^0 + e^+2 = Sn^+2

Answer 21

almost correct :P
The only mistake is that the electrons are on the wrong side of the arrow.
So it compensates the number of electrons on both sides, the way you wrote it, the left side has -2 and the right side has +2, It needs to be 0 on both sides.
So you add the electrons on the right.
\(\Large\sf Sn^0\to Sn^{2+}+2e^-\) this way both sides are 0 because +2 -2 = 0

Answer 22

oh!!!!

Answer 23

So now you have the 2 half reactions:
\(\Large\sf Oxidation:~Sn^0\to Sn^{2+}+2e^-\)
\(\Large\sf Reduction:~Ag^+ +e^- \to Ag^0\)

The only thing left is to balance the number of electrons on both sides, as @Preetha said..
To do this both half reactions should gain and lose the same amount of electrons.
You need to add coefficients to reactants and products so that the oxidation has the same \(\sf +~e^-\) as the reaction. So you multiply one of the reactions.

Which reaction do you multiply with what number to get the same amount of electrons gained and lost? the reduction or the oxidation?

Answer 24

IDK

Answer 25

Ox = \(+2e^-\)
Red= \(+e^-\)

We need the reduction to have +2e- too just like the oxidation does.
how do you multiply the reduction so that it is +2e- instead of +1e-?

Answer 26

multiply by 2

Answer 27

YES!
So what do you get when you multiply \(\sf Ag^+ +e^- \to Ag^0\) by 2?

Answer 28

Just add coefficients to every reactant and product

Answer 29

how

Answer 30

2Ag^+ + 2e^- arrow 2Ag^)

Answer 31

yes!!!!!
So now you have 2 half reaction with the same number of electrons gained and lost.
Now add the 2 reactions. You don't have to write down the electrons because they cancel eachother out.

Answer 32

???

Answer 33

\(\Large\sf Oxidation:~Sn^0\to Sn^{2+}+2e^-\)
\(\Large\sf Reduction:~2Ag^+ +2e^- \to 2Ag^0\)

You add these 2 reaction by putting all elements on the left side of the arrow (so Sn and 2Ag+) on the left side of the overall reaction and you do the same for the right side (Sn^2+ and 2Ag)

\(\Large\sf Overall~reaction:~2Ag^++Sn+2e^-\to Sn^{2+}+2Ag+2e^-\)
You delete the electrons because the cancel eachother out
You'd get:
\(\Large\sf Overall~reaction:~2Ag^++Sn\to Sn^{2+}+2Ag\)

Answer 34

Thx your a great help!!!! God bless

Answer 35

\(\overline{\underline{\LARGE{\color{gold}{\bigstar~}}\Large\tt\color{green}{You’re~Welcome!!}\LARGE{\color{gold}{~\bigstar}}}}\)

If you're having trouble more of these questions i'd be happy to explain them to you \(\Large \ddot\smile\)

Answer 36

awww....thx!!!! Your the best chemistry tutor I evr had!!! U taught me more then my chem. teacher ever had in 1 week!