I'm having trouble balancing these two oxidation-reduction reactions that occur in a basic solution. Can anyone help? Please? The numbers after the formulas are subscripts and any with a minus sign means they are negatively charged (-1). Thanks!!
Al(s)+ MnO4-(aq)---> MnO2(s)+Al(OH)4-(aq)
NO2-(aq)+Al(s)-->NH3(g)+AlO2- (aq)
Here are the two half equations (unbalanced) for the first one.
Al + 4OH^- ==> Al(OH)4^- + 3e
MnO4^- + 2H2O + 3e ==> MnO2 + 4OH^-
Add the two equations to obtain the balanced equation.
Certainly! Balancing oxidation-reduction reactions can sometimes be challenging, but with a systematic approach, we can find the balanced equations.
First, let's balance the oxidation-reduction reaction between Al(s) and MnO4-(aq):
Step 1: Assign oxidation numbers to all elements in the reaction.
The oxidation state of Al(s) is 0 since it is an elemental form.
The oxidation state of Mn in MnO4-(aq) is +7.
The oxidation state of O in MnO4-(aq) is -2.
The overall charge of the ion is -1, so the oxidation state of Mn is +7.
Step 2: Identify the atoms undergoing oxidation and reduction.
In this reaction, Al is being oxidized (its oxidation number increases from 0 to +3), while Mn is being reduced (its oxidation number decreases from +7 to +4).
Step 3: Write separate half-reactions for oxidation and reduction.
Oxidation half-reaction:
Al(s) → Al(OH)4-(aq) (Al changes from 0 to +3)
Reduction half-reaction:
MnO4-(aq) → MnO2(s) (Mn changes from +7 to +4)
Step 4: Balance atoms and charges in each half-reaction.
Oxidation half-reaction:
2Al(s) → 2Al(OH)4-(aq) (balanced the number of Al atoms)
Adding 8OH- (aq) on both sides to balance the charges, we get:
2Al(s) + 8OH-(aq) → 2Al(OH)4-(aq)
Reduction half-reaction:
3MnO4-(aq) → 3MnO2(s) (balanced the number of Mn atoms)
Step 5: Balance the number of electrons transferred in each half-reaction.
To balance the electrons transferred, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
6Al(s) + 24OH-(aq) → 6Al(OH)4-(aq)
3MnO4-(aq) → 3MnO2(s)
Step 6: Combine the two half-reactions.
Since 6 electrons are transferred in the oxidation half-reaction and 3 electrons are transferred in the reduction half-reaction, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 6 to equalize the electrons transferred:
12Al(s) + 48OH-(aq) → 12Al(OH)4-(aq)
18MnO4-(aq) → 18MnO2(s)
Final balanced equation in basic solution:
12Al(s) + 18MnO4-(aq) + 48OH-(aq) → 12Al(OH)4-(aq) + 18MnO2(s)
Now let's balance the oxidation-reduction reaction between NO2-(aq) and Al(s):
Step 1: Assign oxidation numbers to all elements in the reaction.
The oxidation state of Al(s) is 0 since it is an elemental form.
The oxidation state of N in NO2-(aq) is +3.
The oxidation state of O in NO2-(aq) is -2.
Step 2: Identify the atoms undergoing oxidation and reduction.
In this reaction, Al is being oxidized (its oxidation number increases from 0 to +3), while N is being reduced (its oxidation number decreases from +3 to -3).
Step 3: Write separate half-reactions for oxidation and reduction.
Oxidation half-reaction:
Al(s) → AlO2-(aq) (Al changes from 0 to +3)
Reduction half-reaction:
NO2-(aq) → NH3(g) (N changes from +3 to -3)
Step 4: Balance atoms and charges in each half-reaction.
Oxidation half-reaction:
2Al(s) → 2AlO2-(aq) (balanced the number of Al atoms)
Adding 4 OH- (aq) on both sides to balance the charges, we get:
2Al(s) + 4OH-(aq) → 2AlO2-(aq) + 2H2O(l)
Reduction half-reaction:
8NO2-(aq) + 8H2O(l) → 8NH3(g) + 8OH-(aq) (balanced the number of N and O atoms)
Step 5: Balance the number of electrons transferred in each half-reaction.
To balance the electrons transferred, we multiply the oxidation half-reaction by 4 and the reduction half-reaction by 8:
8Al(s) + 16OH-(aq) → 8AlO2-(aq) + 8H2O(l)
8NO2-(aq) + 8H2O(l) → 8NH3(g) + 8OH-(aq)
Step 6: Combine the two half-reactions.
Since 8 electrons are transferred in the oxidation half-reaction and 8 electrons are transferred in the reduction half-reaction, we can combine the two half-reactions directly:
8Al(s) + 8NO2-(aq) + 16OH-(aq) → 8AlO2-(aq) + 8NH3(g) + 8H2O(l)
Final balanced equation in basic solution:
8Al(s) + 8NO2-(aq) + 16OH-(aq) → 8AlO2-(aq) + 8NH3(g) + 8H2O(l)
I hope this helps! If you have any further questions, please let me know.