Half Equations Questions i had troble with (1 Viewer)

[lanvins]

1. 2S2O4 -2(aq) + H20(l) = S2O3 -2(AQ) + 2HSO3 - (aq)

Write the ionic half equations for the oxidation and reductions that occur when sodium dithionite is mixed with water

2. In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed.

Calculate the mass of zinc that would be needed to react completely with 8.g of MnO3 in a dry cell.
I got 6g but thats wrong according to the book

3. The thermite process can be used to weld lengths of railway track together. A mould placed over the ends of the rails to be joined is filled with a charge of aluminium powder and iron(iii) oxide. When the mixture is ignited, a redox reaction occurs to form molten iron, which joins the rails together.
a. Write a healf equation for the conversion of iron(iii) oxide to metallic iron.
i got Fe2O3 + 6H+ +6e- = 2Fe + 302- but thats wrong according to the book

b. Write the overall equation for the thermite process.

 

[Undermyskin]

2. The answer is 3.0 g. (thanks for your typo, I nearly got crazy that all I remembered about dry cell was wrong)

Anyway, here is how I do it: you can include the states on your own.

Zn --> Zn 2+ + 2e-

NH4+ + MnO2 + H2O + e- --> Mn(OH)3 + NH3

n MnO2 = 8/ (54.9+32) = 0.092 mol
---> n e- = 0.092 mol --> n Zn = 0.046 mol

--> m Zn = 0.046 * 65.4 = 3.01 g=3.0 g (2 S.f)

3. b. Al(s) + Fe2O3 (s) --> Al2O3 (s) + Fe (s)

a. Fe2O3 + 6e- --> 2Fe +3O 2-

1. I don't think I understand. Do you make any mistakes (again) when typing the 1st equation? Is dithionite S2O4 2-?

 

[WantToDoBetter]

2. Its been a while since i have done stoci so i couldve got it wrong.

Equation (I THINK) is:
Zn(s) + 2 MnO2(s) + ZnCl2(aq) + 2 H2O(l) → 2 MnO(OH)(s) + 2 Zn(OH)Cl(aq)
Zn=65.5
MnO2=71 and there is 8g of it.

8/87= 0.092 mols of MnO2
mole ratio is 1/2 (Divide by 2)
=0.046 mols of Zn
0.046 times 65.5
=3.0g

Mole ratio strikes again , divide 6 by 2. Always look at the equation.

Let me know if i got it right. (Bah, helps if i could add lol)
Cheers.

 

[lanvins]

i'm still a little confused about Q3 why did you balance it with 02- and how did u know to do this (where did it suggest in the question) ? Also, how did u do Q3 b, like where did Al2O3 come from?

when balancing half equations, in what instances do you use oxygen to balance the equation instead of water and why???

 

[Undermyskin]

No, I should ask you where did you get the H+ from??? Um, I'm not quite sure about the equation since I suppose there shouldn't be any half equation. Since the question asks you to write half equation: iron ions to metallic ions, and since Fe2O3 is insegregatable, you should make sure everything appears on both sides is balanced. That's how you get O 2- on the right hand side. I told you I wasn't sure about the existence of Oxygen as an ion like this but that is it if you have to write a half equation. O 2- should be in with Al. (?) Oops, I'm so sorry. I corrected the equation I wrote already. Hope that makes sense. I forgot to include 3 in front of O 2-. My bad.

In part b, the overall equation is a redox reaction. How do you write this? Simply like Zn + CuSO4 in which Zn replaces Cu in CuSO4.

You have the answer for part a, eh? Why don't you post it up here?

 

[Tim035]

Here are some basic steps for balancing equations involving H2O and H+
After concluding which species is being reduced and which is being oxidised, split ur equation into 2 half equations one involving the oxidized species and the other the reduced species, then proceed as follows:

1- Balance atoms other than O and H
2- Balanced O by adding H2O to one side of the equation
3- Balance H by adding H+ to one side of the equation
4- Balance charge by adding e- to one side of equation
5- Complete this for the 2nd half equation
6- Multiply each half equation by an integer, as required so that: the number of electrons lost in one equation= the number of electrons gained in another.
7- combine the two scaled half equations together and cancel the equal number of species common to both sides.