A level chemistry Redox help please.

In which of the following conversions is the copper reduced?
(i)******[Cu(H2O)6]2+ → [CuCl4]2−
(ii)*****[Cu(H2O)6]2+ → Cu(H2O)4(OH)2
(iii)****Cu → CuCl2
(iv)****[Cu(H2O)6]2+ → CuCl

Reply 1

TypicalNerd

Original post by abbAz

In which of the following conversions is the copper reduced?
(i)******[Cu(H2O)6]2+ → [CuCl4]2−
(ii)*****[Cu(H2O)6]2+ → Cu(H2O)4(OH)2
(iii)****Cu → CuCl2
(iv)****[Cu(H2O)6]2+ → CuCl

Start by defining reduction in terms of how the oxidation number changes.

Reply 2

abbAz

Original post by TypicalNerd

Start by defining reduction in terms of how the oxidation number changes.

The oxidation state lowers when reduced

Actually nevermind, i see where i went wrong. There's another questions about redox that I was struggling with.

Which one of the following contains the metal with the lowest oxidation state?
A CrO2F2
B [Cr2O7] 2 - The 2 is squared not the charge
C [MnCl6] 2−
D [Mn(CN)6] 3−

(edited 5 months ago)

Reply 3

abbAz

Actually nevermind, i see where i went wrong. There's another questions about redox that I was struggling with.

Which one of the following contains the metal with the lowest oxidation state?
CrO2F2
[Cr2O7] 2 - The 2 is squared, not the charge
[MnCl6] 2−
[Mn(CN)6] 3−

Reply 4

TypicalNerd

The sum of the oxidation states of all the elements in a species will always equal the charge it has.

For example, let’s look at MnO4^-:

Let’s call the oxidation state of Mn x. Oxygen in compounds almost always has a -2 oxidation state, so:

Charge = sum of oxidation states
==> -1 = 4(-2) + x
-1 = x - 8
+7 = x

So using this approach, can you work out the oxidation state(s) of Cr in CrO2F2 and Cr2O7^2-and the oxidation state of Mn in [MnCl6]^2-?

You can also use a similar rule with complex ions. The cyanide ion (CN^-) has a charge of -1.