Balancing redox reactions
Balance in acid solution
1. PH3+ I2 -> H3PO4 + I-
2. ClO3- + Cl- -> Cl2
Balance in basic solution
Se -> Se2- + SeO3^2-
BrO- + Cr(OH)4^- -> Br- + CrO4^2-
To balance redox reactions, we need to follow a systematic approach. Let's go step by step for each equation you provided.
Balancing in Acidic Solution:
1. PH3 + I2 -> H3PO4 + I-
Start by assigning oxidation numbers to each element:
P: +5 in H3PO4 (it is bonded with three more oxygens, so each oxygen has an oxidation number of -2 making P +5)
H: +1 in H3PO4 (hydrogen typically has +1 oxidation number)
I: 0 in I2 (element in its elemental form has a zero oxidation number)
Then, identify which elements are oxidized and reduced:
P: goes from -3 (in PH3) to +5 (in H3PO4) - oxidized
I: goes from 0 (in I2) to -1 (in I-) - reduced
To balance the number of atoms for each element, begin by balancing the atoms that are undergoing oxidation or reduction:
P: 1 atom on each side
I: 2 atoms on each side
Next, balance oxygen atoms by adding water (H2O) to the side that contains the element needing more oxygen:
P: 4 O atoms in H3PO4, so add 4 H2O on the reactant side to balance (now there are 12 H atoms on the reactant side)
I: 0 O atoms in I2, no water needed
Then, balance hydrogen atoms by adding hydrogen ions (H+) to the side that requires more hydrogen:
P: 12 H atoms on the reactant side, so add 12 H+ ions on the product side to balance
I: 0 H atoms in I2, no hydrogen ions needed
Now, look for additional charges that need to be balanced. This can be achieved by adding electrons (e-) to the side requiring more negative charge:
P: 12 H+ ions on the product side, so add 12 e- on the same side to balance the charges
I: I- ion has a negative charge, no electrons needed
Finally, check if the charges and atoms are balanced on both sides of the equation:
P: 12 H, 1 P, 16 O, and 12 e- on both sides
I: 2 I and 2 e- on both sides
The balanced equation is:
PH3 + I2 -> H3PO4 + 12 H+ + I- + 12 e-
2. ClO3- + Cl- -> Cl2
Assign oxidation numbers to each element:
Cl: -1 in Cl- (chlorine typically has -1 oxidation number)
O: -2 in ClO3- (the overall charge of ClO3- is -1, so the sum of oxidation numbers must equal -1)
Then, identify which elements are being oxidized and reduced:
Cl: goes from -1 (in Cl-) to 0 (in Cl2) - oxidized
O: already in its highest oxidation state of -2
Balance the number of atoms undergoing oxidation or reduction:
Cl: 1 atom on each side
O: no atoms undergoing a change in oxidation state
Next, balance oxygen atoms by adding water (H2O):
Cl: no oxygen atoms needed
O: 3 O atoms in ClO3-, so add 3 H2O on the product side to balance (now there are 6 H atoms on the product side)
Balance hydrogen atoms by adding hydrogen ions (H+):
Cl: no hydrogen atoms needed
O: 6 H atoms on the product side, so add 6 H+ ions on the reactant side to balance
Now, check for additional charges that need to be balanced:
Cl: no charges to balance
O: ClO3- has a negative charge, so no additional electrons needed
Finally, ensure that the charges and atoms are balanced on both sides of the equation:
Cl: 1 Cl on both sides
O: 6 O, 6 H, and 6 e- on both sides
The balanced equation is:
ClO3- + 6 H+ + 6 Cl- -> Cl2 + 3 H2O + 6 e-
To balance redox reactions in basic solution, we will follow similar steps, but we need to account for the presence of hydroxide ions (OH-) acting as a base. Let's balance the third equation:
Se -> Se2- + SeO3^2-
Assign oxidation numbers to each element:
Se: 0 in Se (element in its elemental form has a zero oxidation number)
Then, identify which elements are being oxidized and reduced:
Se: goes from 0 (in Se) to -2 (in Se2-) - oxidized
Se: goes from 0 (in Se) to +4 (in SeO3^2-) - reduced
Balance the number of atoms undergoing oxidation or reduction:
Se: 1 atom on each side
Se: 1 atom on each side
Next, balance oxygen atoms with water (H2O) as before:
Se: 3 O atoms in SeO3^2-, so add 3 H2O on the product side to balance (now there are 6 H atoms on the product side)
Se: no oxygen atoms needed
Balance hydrogen atoms with hydrogen ions (H+):
Se: 6 H atoms on the product side, so add 6 H+ ions on the reactant side to balance
Se: no hydrogen atoms needed
Now, balance charges with hydroxide ions (OH-):
Se: no charges to balance
Se: -2 charge in Se2-, so add 2 OH- ions on the reactant side to balance
Finally, check if the charges and atoms are balanced on both sides of the equation:
Se: 1 Se, 6 H, and 2 OH- ions on both sides
Se: 1 Se, 3 O, and 6 H on both sides
The balanced equation is:
Se + 6 H2O + 2 OH- -> Se2- + SeO3^2- + 6 H
For the last equation, "BrO- + Cr(OH)4- -> Br- + CrO4^2-", the process is similar. Try balancing it using the steps provided above.