How to balance simple redox reactions.

OXIDATION STATE RULES

Oxidation numbers/states are used to balance redox reactions. Balancing more complicated redox-reactions can be quite challanging. First you must master the rules.

What is the oxidation number of an element uncombined with other elements? (For instance Cl₂) ?

Zero, zipp, nada.

What is the sum of all the oxidation numbers of all the atoms in a molecule/a spieces?

It is equal to the total charge. So for instance, MgO is an uncharged molecule. Since O almost always have the oxidation number (-2) when combined with other spieces (see below) Mg must have the oxidation number (+2) in this case.

What is the oxidation number of H when combined with other spieces?

(+1). Consider the H₂O molecule – O has the oxidation number (-2), hence the two hydrogens must contribute with one (+1) charge each!

What is the oxidation number of elements in group 1 and 2?

It is equal to their group number!!!

What is the oxidation number of the halogens?

(-1) – unless combined with O or another halogen. F is (-1) in all compounds.

What is the oxidation number of oxygen?

It is (-2) in most of its compounds. However, in combination with F which always has the oxidation number (-1). Other exceptions : peroxides (O₂^2-), superoxides and ozonides. These molecules are so uncommon that I doubt you'll have to remember it for now.

What does an increase in the oxidation number/state of a spieces indicate?

Oxidation! (loss of electrons = increased oxidation state)

What does a decrease in the oxidation number/state of a spieces indicate?

Reduction! (gain of electrons)

TEST YOUR SELF :

What is the oxidation number of S in H₂S?

(-2)

What is the oxidation number of P in P₄O₆?

(+3) – because the sum of the oxidation numbers of O₆ is 6 ^x(-2) is -12 and the molecule is uncharged the P₄-part of the molecule must equal 12. 12 divided over 4 molecules is of course +3.

What is the oxidation number of Cl in ClO^-?

(+1)

What is an oxidation agent? What happens to it during a redox-reaction?

An oxidation agent (or oxidant) causes oxidation!!! It makes the other spieces in the reaction go through oxidation because the oxidation agent stels electrons. Hence the oxidation agent (or oxidant) gets reduced in the process!!!

What is a reducing agent (or reductant)? What happens to it during a redox-reaction?

It causes reduction, by giving away its electrons. Hence it gets oxidized in the process.

How can you identify the oxidant and reductant in a redox reaction?

You consider the rules listed above. You then compare the oxidation numbers of the spieces before and after the reaction (on each side of the →). I an element has NOT undergone a change in oxidation number it has not been reduced nor oxidized.

Test your self : consider the following reaction, called the claus process :

2H₂S_(g) + SO_2(g) → 3S_{( s)} + 2H₂O Identify the oxidant and reductant!

S in H₂S_(g) goes from (-2) to (0), so it is oxidized and thus the reducing agent.

S in SO_2(g) goes from (+4) to (0), so it is reduced and thus the oxidation agent.

What is the most important rule to consider when balancing redox reactions?

Electrons cannot be lost or created – so the number of electrons that the reducing agent loses must add up to the number of electrons gained by the oxidant.

Balance the reaction : Cu_(s) + Ag⁺→ Cu²⁺ + Ag_(s)

This reaction appears to be balanced because the number of atoms is the same on both sides of the arrow. However the number of electrons is not the same on both sides!!!

Thus we must multiply the spieces with insufficient number of electrons with the number 2 (in this case) to make it all add up. The balanced reaction is thus :

Cu_(s) + 2Ag⁺→ Cu²⁺ + 2Ag_(s)

Write a balanced redox-reaction from the following skeleton reactions:

NO₂ + O₃ → N₂O₅ + O₂?

2NO₂ + O₃ → N₂O₅ + O₂

S₈ + Na → Na₂S?

S₈ + 16Na → 8Na₂S

Cr²⁺ + Sn⁴⁺ → Cr³⁺ + Sn²⁺ ?

2Cr²⁺ + Sn⁴⁺ → 2Cr³⁺ + Sn²⁺

Ok, so tomorrow I will move on to more complicated reactions, taking place in acidic /basic solutions. Then I'll cover galvanic cells and its relation to the equilibrium constant. Ciao Manhattan for now!

Brief and basic redox-chemistry

I've always hated redox reactions and electro chemistry... It's really not that difficult – at least not more difficult than anything else within the field of chemistry – but it's just so... messy, so much to take into account. And if you forget just one detail, everything falls into bits and pieces. Anyhow, I plan to write some sort of summary or whatever about it in the following days. This is part one, starting with the basics.

What is electricity?

Electrons passing from one metal to another. Thus electrictity may be used in order to make chemical reactions occur or the other way around : we can use chemical reactions to produce electricity!

Electrochemistry can also be used to meassure pKa-values, actitvity in the brain, etc...

What is the definition of an oxidation?

The loss of electrons!

Let's consider the following two reactions :

2Mg_(s) + O_2(g) → 2MgO_(s) The solid MgO consists of two ions; 2Mg²⁺_(s) + 2O^2- _(s)

The reaction above describes the transfer of electrons from Mg to O. Mg loses electrons that go to oxygen instead. When Mg_(s) looses its electrons it is transformed to Mg²⁺ ions, and these ions start a relationship with the 2O^2- ions. Remember that opposites attracts! The resulting solid is a salt.

The reaction below is essentially the same, Mg loses electrons, but to Cl this time. Still, it is called “oxidation” since electrons are lost, even though no oxygen is involved.

2Mg_(s) + Cl_2(g) → 2MgCl_2(s) The solid MgCl₂ consists of two ions; 2Mg²⁺_(s) + 2Cl^-_(s)

How can you determine wether an element has undergone oxidation or not?

Its charge is increased. This rule does often, but not always, apply. It can be used on anions as well as on uncharged speices. Consider this reaction :

2 NaBr_(s) + 2Cl_2(g) →2NaCl + Br_2(l)

Br is oxidized by Cl in this case. One might say that Cl breaks up the marriage between Na and Br, since Na is more attracted to Cl. That's how I understand things. (My chemistry professor would kill me...) The charge in Br thus goes from (-1) to (0). I will go through the basic rules of “oxidation numbers” or “oxidation states” further on.

Define “reduction”!

This refers to the opposite of an oxidation; if an atom is reduced it gains electrons. In the case above, 2 NaBr_(s) + 2Cl_2(g) →2NaCl + Br_2(l), Cl is reduced, because its chage goes from (0) to (-1). It takes electrons! The same rule applies to oxygen in the MgO formation described above. This touches on something rather important; electrons cannot just get lost through oxidation, they are in fact particles that go somewhere. So : whenever oxidation occur, reduction will occur.

Identify the spices that have been oxidized and reduced in the following reaction!

3 Ag⁺_(aq) + Al_(s) → 3Ag(s) + Al³⁺(aq)

Ag⁺ gets reduced. Its charge goes from (1+) to (0), it gains electrons that counteracts the positive charge. Al gets oxidized. It loses electrons and goes from (0) to (+3).

Identify the species that have been oxidized or reduced in the following reaction!

2 Cu⁺ _(aq) + I_2(s) → 2 Cu²⁺ + 2I^-

Well, the charge of Cu⁺ increases from (1+) to (2+), so it is oxidized! The charge of I_2(s) is lowered from (0) to (-1) so it is reduced.

What is oxidation number, what is oxidation state, what's the difference?

These terms are often used in the same context. But to be precise (or nerdy, rather) – an oxidation number is given to a spieces, depending on wether it is in a molecule or not, what kind of molecule it is, etc. The oxidation number corresponds to a condition, denoted by oxidation state.

So Mg²⁺ has the oxidation number (2+) which corresponds to the oxidation state (2+).

In short : no real difference, don't worry your pretty little head.

Eum that's all I have (made...) time for today. More to come. Hope this helps anyone.

Thermodynamics - calculations of ∆H°, ∆S° and K

Howdy folks!

Time for some thermodynamics. I'll go through strategies of solving common questions.

1. How does one calculate ∆H° for the following reaction :

CO_{2 (g, 30 mbar)} + H₂O → H⁺ _{(aq, pH =8)} + HCO^-3_{(aq, c=010 mM)}

First you have to find values corresponding to each of the species in the reaction. I don't know where you find yours, but I use SI Chemical data, table 5. Units: kJ/mol

Reactants:

CO_{2 (g, 30 mbar)} = -394

H₂O = -286

Products :

H⁺ _{(aq, pH =8)} = 0

HCO^-3_{(aq, c=010 mM)} = -690

Once you've listed these values, use the formula : (product+product) – (reactant - reactant)

So, in this case : (0 +(-690 )) - (-394) - (-286) → -10 kJ/mol

∆H° = -10 kJ/mol

In order to find ∆S°, use a similar strategy. First; find the S°_f for the different spieces. Units : J/K/mol

Reactants:

CO_{2 (g, 30 mbar)} = 214

H₂O = 70

Products :

H⁺ _{(aq, pH =8)} = 0

HCO^-3_{(aq, c=010 mM)} = 98

Again, use the formula : (product + product) – (reactant- reactant) = 0 + 98 – 214 – 70 = -186 J/K/mol.

∆S° = -186 J/K/mol.

Ok, moving on to how you can calculate K for such a reaction.

The equation is : ∆G° = ∆H° - T∆S°

We've just calculated the values corresponding to ∆H° and ∆S°, -10 kJ/mol & -186 J/K/mol, respectively.

T has to be tranformed into Kelvins. 25°C corresponds to 298°K . Furthermore, we have to transform -10 kJ/mol into J/mol; -10∙10³ J/mol. Now, just insert your values into the equation

∆G° =- -10∙10³ J/mol – 298 ∙ (-186 J/K/mol) = 45428 J/mol OR 45,4 kJ/mol.

K is related to ∆G° in this way : K = e^(∆G°/RT) R denoting the gas constant; 8,314J/K/mol or 8314 kJ/K/mol. So, in this case : K = e- ^{(45,4/8314∙298)} gives 1,08^-08

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Another one then...

Calculate ∆H° for the following reaction :

2 Fe₂O_{3 (s)} + 3 C _(s) → 4 Fe_(s) + 3 CO_{2 (g)}

2 Fe2O3 (s) = -824

Fe = 0

3 CO₂ = 214

3 C = 0

(Product + product) – (reactant-reactant) = 0 + 3(-394) - 2(-824 – 0) = 466 kJ/mol

So ∆H° = 466 kJ/mol

Calculate ∆S° for the reaction 2 Fe₂O_{3 (s)} + 3 C _(s) → 4 Fe_(s) + 3 CO_{2 (g)}

2 Fe2O3 (s) = 87

Fe = 27

3 CO₂ = 214

3 C = 6

Ja osv...

6. 2011-02-24

Calculate ∆H° and ∆S° at 25°C for the following reaction :

C(s) + 2H₂ (g)→ CH₄

CH₄ = -74

C = 0

H₂ = 0

(product + product) – (reactants – reactants) = -74 – 0 - 2∙ 0 = -74 kJ/mol

Enthalpy : -74 kJ/mol

∆S° is calculated through S°_f -values.

CH₄ = 186

C = 6

H₂ = 131

186 – 6- (2∙131) = -82 J/mol/K

Entrophy : -82 J/mol/K

Calculate K for the reaction!

∆G° = ∆H° - T∆S° -74∙10³ J/mol- (298)∙(-82 J)) = enligt mig : 74000- 298 x -82 = - 4,9862 ∙10⁴

• 4,9862 ∙10⁴ /8,314∙298 = -20 → e20 = 4,8∙10⁸

The autoprotolysis of water explained.

Calculate the molarity of OH^- in solutions by the following H3O⁺ concentrations!

a. 0,02 mol/L H3O⁺ . → 1 · 10^-14 / 0,02 = 5 · 10^-13 M/L OH^-

b. 3,1 mM/L H3O⁺ → 1 · 10^-14 / 0,0031M = 3,22 · 10^-12 M/L OH^-

The calculations above lean on the knowledge of the autoprotolysis of water.

Describe the autoprotolysis of water!

The autoprotolysis of water referes to the fact that the H₂O molecule is amphoprotic – it can operate both as a Broenstedt base and as an acid. Proton transfer occurs in pure water as well as in other aqueous solutions. This means that there will always be an exchange of protons between water molecules with H3O⁺ and OH^- molecules as the result. This reaction (just as any chemical reaction) leads to an equilibrium. The equilibrium constant (K_W) is wiritten as follows :

H₂O ↔ H3O⁺ +OH^- → K_W = [H3O⁺]·[OH^- ]/[H₂O]

The activity of H₂O is close to 1 and hence excluded from the equation. K_W is thus written as :

K_W = [H3O⁺]·[OH^- ] and always equals : 1 · 10^-14 M at 25 º C!

In pure water [H3O⁺] [OH^- ] respectively is...?

1 · 10^-7 , both of them.

Does K_W increase or decrease with temperature?

It increases.

The most important thing to understand about K_W is that it, just as all equilibrium constants, always remains the same! That is, if we increase [H3O⁺] , [OH^- ] will decrease enough to preserve

K_W = 1 · 10^-14 M!!!

Acid and bases - part 1.

Define molecular compunds!

Molecular compounds always contain two non-metals, NH3, CO2 etc. In oher words : they are easy to identify. They do not conduct electricity when dissolved in H2O because they are not formed by two ions of opposite charge.

Define ionic compounds!

They are compounds held together by ionic bonds, formed between ions of opposite charges. You probably all know this, but a good rule of thumb is that the cation is usually a metal!

What defines the Brœnstedt acid/base theory?

It is the classic definition of acids as proton donors and bases as proton acceptors. For instance : H₂SO₄ is an acid. It loses its proton to water and is thus transformed into its conjugate base : HSO4-.

Describe the Lewis-theory!

It is a WIDER definition of acids and bases and goes beyond simple proton transfer. Thus other substances than proton acceptors and donors can be denoted as acids or bases according to the Lewis theory, as it focuses on electron pair donations and acceptations.

A Lewis acid ACCEPTS an electron pair! (ACCEPT and ACID = two A:s!)

A Lewis base accordingly donates an electron pair! For instance a Lewis base may give an electron pair to a proton that accepts it - and a covalent bond is formed.

Every Brœnstedt base is a Lewis base, but not every Lewis base is a Brœnstedt base. For instance : CO is a Lewis base, because it donates an electron pair to certain METALS. But it is NOT a Brœnstedt base, since it does not accept protons.

Many nonmetal oxides are...?

Lewis acids. They react with water and accepts the electron pair of the oxygen in the H2O molecule. The product is a Brœnstedt acid!

What defines an acidic oxide?

As in the example above, it is a molecular oxide that reacts with water to form a Brœnstedt acid. Acidic oxides may also react with Brœnstedt bases, forming H₂O and salt.

Define a basic oxide!

It is a ionic compound. It reacts with water to create Brœnstedt bases. It may also react with acids, thus forming salt and H₂O.

Summary :

Acidic oxides are?

Molecular compunds!

They react with?

H₂O and the result is a Broensted acid as in the example below :

CO₂ + H₂O → H₂CO₃

Or with...?

A base with a salt + H₂O as a result! See example below :

CO₂ + 2NaOH → Na₂CO₃ + H₂O

Basic oxides are?

Ionic compounds!

They react with?

H₂O and the result is a Broensted base as in the example below :

CaO + H₂O → Ca(OH)₂

So now that's that. But whatabout oxides formed by the metalloids?

These elements form amphoteric oxides; that is they can react with both bases and acids.

Examples below :

Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O

or

Al₂O₃ + 2 NaOH + 3H₂O → 2[NaAl(OH)₄]

Several of the d-element also form amphoteric oxides.

Write the two different equilibria between HCO₃^- and H₂O!

1. When HCO₃^- acts as an acid : HCO₃^- + H₂O → CO₃^2- + H3O⁺

2. When HCO₃^- acts as a base : HCO₃^- + H₂O → H₂CO₃ + OH^-

State wether the following oxides are acidic, basic or amphoteric! BaO, SO₃, As₂O₃, Bi₂O₃?

BaO = basic

SO₃ = acidic

As₂O₃ = amphoteric

Bi₂O₃ = basic

The easiest way to solve the question above is of course by looking at the periodic table.

More to come, bitches! Love/A

Tenta-musik

Lugn men ändå hetsig musik.

PCR - what is it and what is the point?

This is a basic introduction to PCR, basically for biochem. noobies. Please do not copy this without asking.

PCR is one of the most essential methods in biochemistry, molecular biology, medicine and forensic science. The procedure can be described as an imitation of the DNA replication that takes place during each mitosis. During PCR all of the enzymes involved, except for the DNApolymerase, have been replaced by heat cycles. Furthermore a PCR buffer, containing two kinds of primers as well as all four nucleotides necessary to build the DNA strands, is necessary to provide an optimal environment for the polymerase. Mg2+ is also an essential ingredient of the buffer since this cation stabilizes the hybridization of primers to the single strands of DNA as well as the bonds between the nucleotides and the DNA polymerase. The DNA polymerase is derived from organisms tolerant to extreme heat, thus able to function even at the high temperatures needed to denature the

DNA molecules. To begin with, the DNA sample of interest is exposed to a temperature of 92°C , thus denaturing the molecule and causing its strands to separate. Then a temperature of approximately 55°C (depending on the lengths of primers and gene fragments) causes the forward and reverse primers to hybridize to one single strand each. The free OH - -ends of the primers allow the DNA polymerase to find its way to the template and begin to build a complementary strand. The whole routine is repeated approximately thirty times, causing the numbers of DNA molecules to increase exponentially and resulting in roughly 1 million copies in only 30 minutes. In short; one of the main advantages of the PCR method is that it provides amplification of the gene of interest, thus ensuring that scientists have endless amounts of the material they want to study.

Kary Mullis won the Nobel Price in 1993 for inventing the method. Some argued at the time that the method is too "simple" for its inventor to earn this honour. However, the incredible advantages that the method has provided to science cannot be underestimated.