Question

For those reactions that are redox, indicate which elements are oxidized. P₄(s)+10HClO(aq)+6H₂O(l)Br₂(l)+2K(s)CH₃CH₂OH(l)+3O₂(g)
$\longrightarrow$ 4H₃PO₄(aq)+10HCl(aq)KBr(s)3H₂O(l)+2CO₂(g). Drag the appropriate items to their respective bins.

Answer 1

Answer:

Explanation:

Hello, at first, the determination of the oxidation states should be attained as follows:

 $P^{0}_{4}+10H^{+1} Cl^{+1} O^{-2} +6H^{+1}_{2}O^{-2}+Br^{0}_{2}+2K^{0}+C^{-4}H^{+1}_{3}C^{-4}H^{+1}_{2}O^{-2}H^{+1} -->H^{+1}_{3}P^{+5}O^{-2}_{4}+10H^{+1}Cl^{-1}+K^{+1}Br^{-1}+H^{+1}_{2}O^{-2}+C^{+4}O^{-2}_{2}$

Now, the following elements exhibit an increase in the oxidation state:

$P^{0}-->P^{+5}\\K^{0}-->K^{+1}\\C^{-4}-->C^{+4}$

The rest of the elements remained with no change in the oxidation state (H and O) or were reduced (Cl and Br).

Best regards.

Answer 2

Answer:

The following elements are oxidized because they lose electron and their oxidation number increased

P4 oxidation number increased from 0 to +5(phosphorus is oxidized)

K oxidation number increased from  0 to +1(potassium is oxidized)

C oxidation number increased from -2 to +4(carbon is oxidized)

Explanation:

This are the redox reaction

P4(s) + 10HClO(aq) + 6H2O(l) → 4H3PO4(aq) + 10HCl(aq)

Br2(l) + 2K(s) → 2KBr(s)

CH3CH2OH(l) + 3O2(g) → 3H2O(l) + 2CO2(g)

Oxidation-reduction reaction, electrons are transferred between two species. Electron are usually lose by one atom and gained by another atom.

Oxidation occurs when an element loses electron and the oxidation number increases while reduction occurs when an element gains electron and the oxidation number decreases.

The element that is oxidized is the element that loses electron but the oxidation number increases.

P4 oxidation number increased from 0 to +5

The oxidation number of phosphorus can be calculated in the product side as follows : 4H3PO4  +1 × 12 + 4x + -8 × 4 = 0

12 + 4x - 32 = 0

4x = 20

x = 20/4 = +5

K oxidation number increased from  0 to +1.

The oxidation number of potassium on the product side can be computed as follows 2KBr

2x + -2 = 0

2x = 2

x = +1

C oxidation number increased from -2 to +4.

Reactant side CH3CH2OH

2x + 6 - 2 = 0

2x + 4 = 0

2x = -4

x = -2

product side  2CO2

2x + 2 × -4 = 0

2x - 8 = 0

2x = 8

x = +4

Note the oxidation number for all single element is 0. The oxidation number for oxygen is -2 and hydrogen is mostly +1