1) 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
Express the oxidation state numerically (e.g., +1).
(ii) Which element is reduced in this reaction?
2HCl+ 2KMnO4 + 3H2C2O4 ? 6CO2 +2MnO2 + 2KCl + 4H2O
Enter the elemental symbol.
O = -2
and the charge of the ion is -2
2xC + (-2)x4 = -2
C = +3
3) Mn is reduced from +7 in KMnO4 to +4 in MnO4.
for KMnO4, it is neutral
K = +1, O = -2
So 1 + Mn + (-2) x 4 = 0 ===> Mn = +7
for MnO2, it is neutral,
O = -2, So Mn + 2 x (-2) = 0 ===> Mn = +4
while Mn is reduced, element C is oxidized.
for H2C2O4, it is neutral
H = +1, O = -2
2x1 + 2xC + 4x(-2) = 0 ===> C = +3
for CO2, it is neutral,
O = -2, So C + 2 x(-2) = 0 ===> C = +4
the oxidation state of C changes from +3 in H2C2O4 to +4 in CO2,
thus C is oxidized.