Answer:
1. Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)
2. CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)
3. Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-
4. CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O
Explanation:
The given equations are redox reaction equations expressed as as half reactions.
The first step is to identify whether the half-reaction is oxidation reduction.
Then the number of electrons gained or lost are added on the right side of the equation.
Appropriate H+ ions and water molecules are added where necessary since the reaction takes place in acidic environment
The atoms of elements involved in the reaction are balanced by adding the correct coefficients.
1. (acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)
The half-reaction is reduction as the oxidation number of chromium changes from +6 to +3. Two Cr⁶+ ions accepts 3 electrons each to form Cr³+ ions
Cr₂O₇²−(aq) + 6e- ---->⟶2 Cr³+(aq)
Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)
2. (acidic) CrO₄²− (aq)⟶---> Cr(OH)₄ −(aq)
The half-reaction is a reduction. One Cr⁶+ accepts 3 electrons to become Cr³+
CrO₄²− (aq)⟶+ 3e- ---> Cr(OH)₄ −(aq)
CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)
3, (acidic) Bi³+ (aq)⟶---> BiO₃− (aq)
The half-reaction is an oxidation. One Bi³+ ion gives up two electrons to become Bi⁵+
Bi³+ (aq)⟶---> BiO₃− (aq) + 2e-
Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-
4. (acidic) CIO −(aq)⟶---> Cl −(aq)
The half-reaction is a reduction. One Cl+ ion accepts two electrons to become Cl- ion.
CIO −(aq) + 2e-⟶---> Cl −(aq)
CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O
