Which atom has a change in oxidation number of –3 in the following redox reaction? K2Cr2O7 + H2O + S - - > KOH + Cr2O3 + SO2
K
Cr***
O
S
right
In the given redox reaction, the atom that has a change in oxidation number of -3 is the chromium atom (Cr).
To determine which atom has a change in oxidation number of -3 in the given redox reaction, we need to examine the oxidation states (oxidation numbers) of each atom before and after the reaction.
In the reactant side (left-hand side):
- The oxidation state of potassium (K) in K2Cr2O7 is +1.
- The oxidation state of chromium (Cr) in K2Cr2O7 is +6.
- The oxidation state of oxygen (O) in K2Cr2O7 is -2.
- The oxidation state of sulfur (S) in S is -2.
In the product side (right-hand side):
- The oxidation state of potassium (K) in KOH is +1.
- The oxidation state of chromium (Cr) in Cr2O3 is +3.
- The oxidation state of oxygen (O) in Cr2O3 is -2.
- The oxidation state of sulfur (S) in SO2 is +4.
Now, let's calculate the change in oxidation number for each atom by subtracting the oxidation number in the reactant side from the oxidation number in the product side:
- K: +1 - (+1) = 0 (no change)
- Cr: +3 - (+6) = -3 (change of -3)
- O: -2 - (-2) = 0 (no change)
- S: +4 - (-2) = +6 (change of +6)
Therefore, the atom with a change in oxidation number of -3 in the given redox reaction is chromium (Cr).
(Redn half-rxn; Cr2O7^2- is Oxidizing Agent)
16HOH + 2K^+ + 4Cr2O7^2- + 12e^- => 4Cr2O3 + 20OH^-
(Cr^+6 => Cr^+3 + 3e^-)
(Oxidation half-rxn; S^o is the Reducing Agent)
12OH^- + 3S^o => 2K^ + 3SO2 + 6HOH + 12e^-
(S^o => S^+4 + 4e^-)
Net Oxdn-Redn Rxn (K^+ = Spectator Ion):
4Cr2O7^-2 + 3S^o + 10HOH => 4Cr2O3 + 3SO2 + 20OH^-
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