Balance the following basic Redox reaction:
Cr3+ (aq) + MnO2 (s) ----------> Mn2+ (aq) + CrO4 2- (aq)
Cr in Cr+3 has oxidation state of +3
Cr in CrO4-2 has oxidation state of +6
So, Cr in Cr+3 is oxidised to CrO4-2
Mn in MnO2 has oxidation state of +4
Mn in Mn+2 has oxidation state of +2
So, Mn in MnO2 is reduced to Mn+2
Reduction half cell:
MnO2 + 2e- --> Mn+2
Oxidation half cell:
Cr+3 --> CrO4-2 + 3e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
3 MnO2 + 6e- --> 3 Mn+2
Oxidation half cell:
2 Cr+3 --> 2 CrO4-2 + 6e-
Lets combine both the reactions.
3 MnO2 + 2 Cr+3 --> 3 Mn+2 + 2 CrO4-2
Balance Oxygen by adding water
3 MnO2 + 2 Cr+3 + 2 H2O --> 3 Mn+2 + 2 CrO4-2
Balance Hydrogen by adding H+
3 MnO2 + 2 Cr+3 + 2 H2O --> 3 Mn+2 + 2 CrO4-2 + 4 H+
Add equal number of OH- on both sides as the number of H+
3 MnO2 + 2 Cr+3 + 2 H2O + 4 OH- --> 3 Mn+2 + 2 CrO4-2 + 4 H+ + 4 OH-
Combine H+ and OH- to form water
3 MnO2 + 2 Cr+3 + 2 H2O + 4 OH- --> 3 Mn+2 + 2 CrO4-2 + 4 H2O
Remove common H2O from both sides
Balanced Eqn is
3 MnO2 + 2 Cr+3 + 4 OH- --> 3 Mn+2 + 2 CrO4-2 + 2 H2O
This is balanced chemical equation in basic medium
Answer:
3 MnO2 + 2 Cr3+ + 4 OH- -> 3 Mn2+ + 2 CrO42- + 2 H2O
Balance the following basic Redox reaction: Cr3+ (aq) + MnO2 (s) ----------> Mn2+ (aq) + CrO4 2-...
Balance each of the following RedOx reactions occurring in basic conditions: Mn2+(aq) + MnO41-(aq) → MnO2(s) Cl2(g) → ClO31-(aq) + Cl1-(aq)
Balance the following redox reaction in acidic solution.... Mn2+(aq)+Zn2+(aq)=MnO2(s)+Zn(s)
Complete and balance the following half-reaction in basic solution Mn2+(aq) → MnO2(s)
Balance the following redox equations by the half-reaction method: (a) Mn2+ + H2O2 → MnO2 + H2O (in basic solution) (b) Bi(OH)3 + SnO22− → SnO32− + Bi (in basic solution) (c) Cr2O72− + C2O42− → Cr3+ + CO2 (in acidic solution) (d) ClO3− + Cl− → Cl2 + ClO2 (in acidic solution) (e) Mn2+ + BiO3− → Bi3+ + MnO4− (in acidic solution)
Complete and balance the following redox reaction in basic solution MnO4-(aq) + Br-(aq) → MnO2(s) + BrO3-(aq) Attempts remaining: 3 Complete and balance the following redox reaction in basic solution 103-(aq) + Re(s) → Re04 (aq) + 10-(aq) Attempts remaining: 3
Balance the following REDOX reaction in basic solution: I2(s) + MnO2(s) ---> I-(aq) + MnO4-(aq) (E0 = -0.060 V) The equilibrium constant for this reaction at 25 degrees C is: 8.30 X 10-7 b. 1.38 X 10-7 c. 0.36 d. 8.30 X 107
Balance the following redox reaction equation first in acidic solution and then in a basic solution: Cr (s) + MnO2 (s) --> Mn+2 (aq) + Cr+3 (aq)
[15] Balance the following redox reactions: {3 points each} a) Cr2O72-(aq) + Cl-(aq) → Cr3+(aq) + Cl2(g) (acidic medium) b) Al(s) + MnO4-(aq) → MnO2(s) + Al(OH)4-(aq) (basic medium)
1. Balance the following reaction in acidic solution. I-(aq) + MnO4-(aq) Mn2+(aq) + I2(s) 2. Write a balanced net ionic equation for the following reaction in basic solution using both methods for balancing a redox reaction: Zn(s) + NO3-(aq) à NH3 (aq) + Zn(OH)42- (aq) 3. Write a balanced net ionic equation for the following reaction in basic solution using both methods for balancing a redox reaction: MnO4-(aq) + C2H5OH(aq) à Mn2+(aq) + HC2H3O2 (aq)
BALANCE the redox equation in a BASIC solution and FIND the oxidizing and reducing agents: CrI3(s) + H2O2(aq) = CrO4^2-(aq) + IO4-(aq) The equation is in an acidic solition