What are the names of these complex ions? And how do you balance this redox equation?

Question

This is really two separate questions, so even if you just answer one that would be great.

And how do you know? What is the method used to name them?

a) [Ni(NH3)6]^2+
b) MnS
c) Cr(OH)3
d) Fe(SCN)^2+


Balance the following redox equation:

I^- + SO4^2- + H^+ --> I2 + H2S + H2O

I seem to remember that you have to do something special with redox reactions when balancing the equations. Do you have to balance the charges too or somthing? Or just balance it like a normal equation?

Answer 1

I don't see why you can't balance it like a normal reaction. Count up the charges and see if they're equal for both sides. Oh, well, I guess the key difference here is that you balance not coefficients but charges.
for the names:
1) Nickel (II) Ammonium
2) Magnesium Sulfide
3) Chromium (III) hydroxide
4) Iron(III) Thiocyanate
Make sure you know the rules of why these are the names:
For #4, for example, you have SCN- with a Fe which yields a 2+ charge. So, Fe must be 3+. Therefore, the ion is Fe (III).

Answer 2

I would first rewrite the equation into:

HI + H2SO4 --> I2 + H2S

In redox equations H2O is used to balance the number of Oxygens and is usually not included in the beginning, but it will work out later.

Iodine undergoes oxidation, as its oxidation number goes from 1- to 0 (I- --> I2).
Sulfur undergoes reduction, as its oxidation number goes from 6+ to 2- (SO4^2- --> H2S).

So balance each oxidation/reduction half reactions separately:

HI --> I2

First balance all elements other than H and O.
2HI --> I2

Then add H2O to balance Oxygen.(not necessary here but is necessary for the other half reaction.)
2HI --> I2

Then add H+ to balance the Hydrogen.
2HI --> I2 + 2H+

Then add e-(electrons) to balance the charges.
2HI --> I2 + 2e- + 2H+

For the other half reaction, follow the same steps above:

H2SO4 --> H2S
H2SO4 --> H2S + 4H2O
8H+ + H2SO4 --> H2S + 4H2O
8e- + 8H+ + H2SO4 --> H2S + 4H2O

so we have two balanced half reactions:
2HI --> I2 + 2e- + 2H+
8e- + 8H+ + H2SO4 --> H2S + 4H2O

multiply the top half reaction by 4 and add the two reactions, cancelling out terms that are the same on both sides of the equation. Multiplying by 4 gives same number of e- on both sides(thus cancelling out), as it is the aim in other redox equations. However in this case, H+ also conviniently cancels out.

Final equation:

8HI + H2SO4 --> H2S + 4H2O + 4I2

expanding it out to fit your original:

8I- + SO4^2- + 10H+ --> 4I2 + H2S + 4H2O

Be sure to check that both sides contain the same number of each elements and equal charges.

Answer 3

a)hexaaminenickel(II)
b)mangenese sulfide
c)chrum(III) hidroxide
d)thoocyanateifon(III)
8I- +SO4<2-> + 10H+ -->4I2 + H2S+4 H2O

Answer 4

a) cation nickelhexamine
B) sulfur of manganese
C) chromium hydroxide
D) iron sulfocyanate

excuse me for the other, I am not sure

Answer 5

Hey sam i did try to look at your question! but i do not know how to solve it =( i tried even before you comment on mine! good luck

Answer 6

i do believe we are in the same lab and have the same idea.

Answer 7

I do not know Ching Ching 2points