Question

In an ion, the sum of the oxidation states is equal to the overall ionic charge. Note that the sign of the oxidation states and the number of atoms associated with each oxidation state must be considered. In OH-, for example, the oxygen atom has an oxidation state of 2 and the hydrogen atom has an oxidation state of +1, for a total of ( -2 ) + ( +1 ) = -1

(i) What is the oxidation state of each individual carbon atom in C2O4 2^-
Express the oxidation state numerically (e.g., +1).



(ii) Which element is reduced in this reaction?
2HCl+ 2KMnO4 + 3H2C2O4 ? 6CO2 +2MnO2 + 2KCl + 4H2O

Answer 1

(i) +3.

(ii) Mn.

Step-by-step explanation:

Hello,

(i)

Based on the diatomic ion, we solve for the unknown oxidation state of carbon, assuming that the oxygen works with -2, by a change balance:

`(C_2O_4)^(-2)\\(C_2^(x) O_4^(-2) )^(-2)\\\\2x-8=-2\\x=+3`

(ii)

We first must each ion's oxidation states as follows:

`2H^(+) Cl^(-)+ 2K^(+)Mn^(+7)O^(-2)_4 + 3H_2^(+)C_2^(+3)O_4^(-2) --> 6C^(+4)O^(-2)_2 +2Mn^(+4)O_2^(-2) + 2K^(+)Cl^(-) + 4H_2^(+)O^(-2)`

Then, for the reduction half reaction we identify the manganese as the element decreasing its oxidation state based on:

`Mn^(+7) +3e^(-)-->Mn^(+4)`

In addition, the carbon is oxidized from +3 to +4 (increase the oxidation state).

Kind regards.