Module 2 - Balancing Redox - Oxidation Number Method - 1
Summary
TLDRThe video script discusses balancing redox reactions, emphasizing the importance of considering charges on atoms and substances. It outlines two methods for balancing: the oxidation number method and the half-reaction method. The script provides a detailed step-by-step guide on how to balance redox equations, including assigning oxidation numbers, identifying oxidation and reduction, determining changes in oxidation numbers, balancing electrons, atoms, and charges. An example is used to illustrate the process, showing how to balance a reaction involving iron and permanganate ions.
Takeaways
- 🔬 Balancing redox reactions involves considering the charges of atoms or substances, which is different from balancing neutral equations.
- 🌐 The law of conservation of mass must be obeyed, along with balancing the number of elements and electrons in the reaction.
- 📐 There are two methods for balancing redox equations: the oxidation number method and the half-reaction method.
- 📖 The oxidation number method tracks changes in oxidation numbers of elements involved in the reaction.
- 📝 Assign oxidation numbers to each atom in the equation as the first step in balancing redox equations.
- 🔍 Identify which substances undergo oxidation and reduction by observing changes in oxidation numbers.
- 🔄 Determine the changes in oxidation numbers for elements involved in the reaction.
- ⚖️ Balance the electrons by adjusting coefficients to ensure the number of electrons lost is equal to the number gained.
- 🌐 Balance the atoms and charges by adjusting coefficients and possibly adding substances like water to the reaction.
- 💧 Adding water to a reaction can help balance the number of oxygen and hydrogen atoms.
- 🔋 Ensure that both the number of atoms and the total charges are balanced in the final equation.
Q & A
Why is it important to consider charges when balancing redox reactions?
-Charges are important in redox reactions because they indicate the transfer of electrons, which is the essence of redox processes. Balancing charges ensures that the law of conservation of charge is upheld, meaning the total charge on both sides of the reaction must be equal.
What are the two methods mentioned for balancing redox reactions?
-The two methods for balancing redox reactions are the oxidation number method (also known as the oxidation statement method) and the half-reaction method (also known as the ion-electron method).
How does the oxidation number method work?
-The oxidation number method works by tracking changes in the oxidation numbers of the elements involved in the reaction. It involves assigning oxidation numbers to each atom, identifying substances undergoing oxidation and reduction, determining the change in oxidation numbers, and then balancing the electrons and atoms.
What is the first step in balancing a redox reaction using the oxidation number method?
-The first step in balancing a redox reaction using the oxidation number method is to assign oxidation numbers to each atom in the equation.
How do you identify which substance undergoes oxidation and which undergoes reduction?
-You identify oxidation and reduction by observing changes in oxidation numbers. A substance that loses electrons (becomes more positive) undergoes oxidation, while a substance that gains electrons (becomes less positive) undergoes reduction.
What does it mean for a substance to be the oxidizing agent or the reducing agent?
-A substance is the oxidizing agent if it causes another substance to be oxidized (loses electrons). Conversely, a substance is the reducing agent if it causes another substance to be reduced (gains electrons).
How do you balance the number of electrons in a redox reaction?
-To balance the number of electrons, you equalize the total increase in oxidation numbers with the total decrease. This is done by adjusting the coefficients of the reactants and products to ensure the number of electrons lost by the oxidized substance equals the number gained by the reduced substance.
Why is it necessary to balance atoms and charges after balancing the electrons?
-After balancing the electrons, it's necessary to balance atoms and charges to ensure that the law of conservation of mass is upheld. This involves adjusting coefficients and possibly adding substances like water to balance the number of atoms and ensure that the total charge on both sides of the equation is equal.
What role does water play in balancing redox reactions?
-Water can be added to a redox reaction to balance the number of oxygen and hydrogen atoms. It also helps in balancing the charges by providing additional protons (H+) or hydroxide ions (OH-) as needed.
Can you provide an example of how to balance a redox reaction involving iron and permanganate ions?
-In the given script, iron (Fe) goes from an oxidation state of +2 to +3, and manganese (Mn) goes from +7 to +2. The changes in oxidation numbers are +1 for iron and +5 for manganese. By balancing the electrons, the coefficients for iron and manganese are adjusted to 5 and 1, respectively. Then, water is added to balance the oxygen atoms, and additional hydrogen ions are added to balance the charges, resulting in a balanced redox reaction.
Outlines
🔬 Balancing Redox Reactions
This paragraph introduces the concept of balancing redox reactions, which involve the transfer of electrons and are not neutral. It explains that in addition to the law of conservation of mass, one must also balance the number of electrons. Two methods are mentioned for balancing redox reactions: the oxidation number method and the half-reaction method. The oxidation number method involves tracking changes in oxidation numbers of elements involved in the reaction. The steps include assigning oxidation numbers to each atom, identifying oxidation and reduction, determining changes in oxidation numbers, balancing electrons, and balancing atoms and charges. An example is given where the oxidation numbers of iron and permanganate are calculated before and after the reaction.
📚 Identifying Oxidation and Reduction
The paragraph discusses how to identify which substances undergo oxidation and reduction in a redox reaction. It uses the example of iron (Fe) and permanganate (MnO4-) to illustrate the process. Iron's oxidation state increases from +2 to +3, indicating it releases an electron and is being oxidized, thus acting as a reducing agent. Conversely, permanganate's oxidation state decreases from +7 to +2, indicating it gains an electron and is being reduced, thus acting as an oxidizing agent. The paragraph also explains how to determine the changes in oxidation numbers and balance the electrons by adjusting the coefficients of the reactants.
🌐 Balancing Atoms and Charges
This paragraph focuses on the final steps of balancing a redox reaction, which include balancing the number of atoms and charges. Using the example from the previous paragraph, it shows how to add water (H2O) to balance the oxygen atoms and how to add hydrogen ions (H+) to balance the hydrogen atoms. The charges are then balanced by ensuring that the total positive and negative charges on both sides of the reaction are equal. The final balanced redox reaction is presented, showing the reactants and products with their respective charges, demonstrating that both atoms and charges are balanced.
Mindmap
Keywords
💡Balancing Redux
💡Charges
💡Redox Reaction
💡Oxidation Number
💡Oxidation
💡Reduction
💡Half-Reaction Method
💡Ion-Electron Method
💡Permanganate
💡Law of Conservation of Mass
💡Balancing Electrons
Highlights
The necessity to consider charges in balancing redox reactions.
Redox reactions involve the transfer of electrons, affecting the balance of chemical equations.
Two methods for balancing redox reactions: oxidation number method and half-reaction method.
Oxidation number method tracks changes in oxidation numbers of elements involved in the reaction.
Assign oxidation numbers to each atom in the equation for balancing redox equations.
Identify substances undergoing oxidation and reduction based on changes in oxidation numbers.
Determine the change in oxidation numbers for elements involved in the reaction.
Balancing the electrons involves adjusting coefficients to match changes in oxidation numbers.
Balancing atoms and charges requires adjusting for the presence of ions and their charges.
The example provided illustrates the step-by-step process of balancing a redox reaction.
The importance of balancing both the number of atoms and the charges in a redox reaction.
The role of water in balancing the oxygen atoms in a redox reaction.
The inclusion of hydrogen atoms to balance the charges in the reaction.
The final balanced redox reaction, showing the transfer of electrons and charge balance.
The law of conservation of mass must be obeyed in addition to balancing elements and electrons.
The significance of the oxidation number method in balancing redox reactions involving charged species.
The practical application of balancing redox reactions in chemistry education.
Transcripts
Now let's proceed to balancing Redux
reaction when you have um an example Uh
the first examples you notice that the
substances or the atoms are not neutral
you observe that the atom or the
substance has charges The question is
will those charges affect
Uh the balancing of
the reaction Okay so at this point We
will now consider the charges of these
atoms or substances because usually when
you were um in junior high school or
senior high school you balance equations
that are Uh neutral meaning They don't
have charges but at this point we have
to consider it now ah
at this point We will now um call it as
a redox reaction whenever we balance the
chemical equation We must obey Again the
law of conservation of mass in addition
to balancing the number of elements in
the chemical reaction one must also
consider balancing the number of
electrons in the in the process and
there are Two methods for balancing the
redox re equations So it is the
oxidation number method also known as
oxidation statement method and the half
reaction method also known as the um ion
electron method so in the oxidation
um number
method It is a technique used to balance
Redux reactions which are chemical
reactions involving the transfer of
electrons Now it works by tracking the
changes in oxidation numbers of the
elements involved in the reaction and
there are some steps that you can
utilize in order to obtain the balance
um redox equation so first is Uh You
have to assign the oxidation number to
each atom in the equation so again we
have to assign the oxidation number to
balance the Uh atoms Okay then we have
to identify which substance undergo
oxidation and which substance undergo
reduction after Identifying the
oxidation and reduction we have to Uh
determine the change in the oxidation
number and after we identify the the
change of oxidation number we have to
balance the
electrons Then after such we have to
balance the atoms and charges so let's
have um some examples so first Uh
considered the unbalance Redux reaction
Okay so take note that it is not your
usual chemical equations where you have
neutral substances or neutral atoms in
this example you notice that each or I
think all of the atoms or substances
have Uh charges So how to balance this
Okay based on what I've discussed You
have to first identify the oxidation
number to each atom in the equation so
we have to identify the oxidation number
of each Okay so first is the ion ion is
an ion and it has a charge of positive
two therefore it has an oxidation number
of pos 2 next is you have permanganate
Okay so the permanganate ah you have
manganese and you have oxygen Okay so
the oxygen again rule number seven has
an oxidation number of
negative and for manganese if you
compute this one it has an oxidation
number of solid compute mn
plus 4 Oxygen is equ to -1 by
substitution for
-2
mn -1 So this is -8
mn -1 So this is
mn -1
transpose yung ne 8 magiging pos 8 so
you have pos 7 so therefore the
manganese and oxidation number of pos 7
now after the reaction the Iron now has
an oxidation number of pos
3 and then the manganese now has
oxidation number
positive Okay so at this point we were
able to identify the number each element
in the equation second is we have to
identify which undergo oxidation and
reduction so notice that the ion mula sa
pos 2 naging pos 3 it becomes more
positive the reason for that is it
releases electron since the ion releases
electron it undergo oxidation process
this
oxidation Okay so meaning it is the one
being oxidized and at the same time it
is
Uh the reducing agent or the reductant
on the other hand if you notice that the
Mangan mula sa pos 7 it now becomes pos
2 The Reason mula sa pos 7 to pos 2 is
humaba ang kanyang Ah yung kany naging
less positive siya because it
since it receives electron it undergoes
a reduction
process TH the permanganate or mno4 is
our is being oxidized and
Our oxidizing Agent Third is we have to
determine the changes in oxidation
number so si ion
ah mula sa
positive 2 naging pos 3 So ang change
niya ay 1 lang Okay tapos si manganese
naman mula sa pos 7 naging
positive 2 therefore ang kanyang change
ay pos
5 Okay so now we were able to determine
the oxidation number the change of
oxidation number of the manganese and
the Iron Okay so next we have to balance
the electrons so notice that the change
of electron is not the same okay ang isa
positive Ah isa one Ay yung isa ay 5
para ma-balance iyan Okay gagawin natin
' ba si
ano si ion ay ay isa pero si manganese
ay lima Okay so paano ba natin ito
ma-balance
what we can do is ' ba si Iron ay isa
ang kanyang kulang Okay so therefore
maglalagay tayo ng one ah one ang
kanyang change So maglalagay tayo ng one
Dian sa mga Mangan coefficient na one si
Mangan naman ay five un change Okay
therefore lalag natin ng five
coefficient si
Iron
okay
okay So ngayon we were able to balance
the number of
electrons third ah fifth and last we
have to balance the number of atoms and
charges so notice na si ion
Okay si IR
okay ngayon Meron ka ng five electron ah
five sa reactant and five sa product si
manganese dito merong isa sa reactant at
meron din Isa siya sa ah sa
product Okay pagdating dito kay
oxygen Okay merong apat sa reactant pero
wala sa product
therefore we have to balance Okay notice
na para ma-balance siya magdadagdag tayo
ng
water
H2O para mabalanse sila
therefore Okay kung titingnan natin ang
water ha Titingnan natin ang water Meron
yang dalawang hydrogen at merong apat ah
Meron yang dalawang hydrogen at Meron
yang isang oxygen Okay pero ang kulang
natin ay
AP na oxygen so therefore dahil apat ang
kulang natin na oxygen so Dapat apatin
dito Okay so apatin dito therefore
maglalagay tayo dito ng apat na water
dito Okay para ma-balance ang oxygen So
kung titingan natin ngayon Apat ang
oxygen sa reactant apat din ang sa
product so Balance na ang oxy
however notice na when you add water you
also include aside from oxygen you also
include
hydrogen Ilan ang kang hydrogen
dito 2 4 so 8 so therefore merong 8 dito
sa ano merong 8 dito
sa tawag nito dito sa product pero
walang hydrogen sa
reactant magdadagdag tayo diyan ng 8 so
magdadagdag tayo dito ng 8 na hydrogen
pero that is an ion Okay so para
ma-balance na iyung number of atom and
charges ah atoms so Balance na iyung
atom pero ngayon How about the
charges is this balance so Tingan natin
Okay so you have here ' ba you have ano
ngayun eight hydrogen
plus lima nito tapos isa Dian or kahit
wala na siguro
yan Tapos merong 5 dito tapos 1 Then you
have
4 water Okay Let us check if they are
balanced
Okay let's check muna kung balance ang
mga um elements So you have
hydrogen
ah iron you have manganese and oxygen
kung balance
pa 8 na
hydrogen 2 * 4 you have 8 iron dito ay
five Dito naman ay five habat manganese
dito one Ito naman ay one one lang dito
ha Ito ay charge Hindi siya number of
Element ha charge lang yan dito ay apat
na oxygen apat
diyan reactant and one ' ba may one yan
dito 1 * 4 so 4 so Balance na ang
atoms How about sa
charges icompute natin
Okay so dito ang charge nito ay 1 ' ba
postive 1 yan Tim 8 is + 8
Okay
2 Tim 5 is 10
posi
1 Tim 1 is - 1 Okay so therefore 8 + 10
is 18 plus - 1 you have pos
17 and Dito naman 3 5 you have
15 have 2 * 1 you have 2 both are
positive so ito ay walang charge so zero
lang Ian therefore 15 + 2 plus 0 is ah
pos 17 so ito ay positive 17 sa reactant
sa product din ay pos 17 therefore the
charges are balan thus our balance red
redox reaction will be look like this
8 h
plus plus 5
ion 2 plus plus
permanganate forming 5 ion 3 plus plus
manganese 2 plus plus water So this is
now Your balance red reaction using
oxidation number me
浏览更多相关视频
Penyetaraan Reaksi Redoks Metode Bilangan Oksidasi | Kimia SMA | Tetty Afianti
Introduction to Oxidation Reduction (Redox) Reactions
The Oxidation Reduction Question that Tricks Everyone!
Oxidizing Agents and Reducing Agents
REAKSI REDOKS - SIMPLE KONSEP - KIMIA (Kursus Online Rp8.000 per BULAN : cek deskripsi)
KONSEP DASAR REAKSI REDOKS
5.0 / 5 (0 votes)