Lecture 50 honors

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Lecture 50honors
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Galvani biography

• Napoleon took power during this time and Galvani
  refused to swear allegiance Galvani was kicked
  out of is position in the scientific
  associations, his wife died and then he died
  dejected and frustrated.
• Volta did not get involved

He got along with everybody
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Check homework

• Go over oxidation numbers number 7 page 611 and
  number 9 which is a redox

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New homework Read pages 612 618do problems
1-10 page 615Finish handout 50
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So as the elements combine and recombine the
electrons shift from one compound to another

• This transportation of electrons form one place
  to another is considered an electric current.
  For some elements this transference of the
  electrons is slow while for others compounds it
  is fast

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Not all redox reactions are easily
balanced.Sometimes you have to balance the
individual elements along with the electrons
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To balance a Redox reactoin you need to follow 6
steps

• Separate the ½ reactions
• Balance the elements (except Oxygen and Hydrogen
  because the rxn is in H2O
• Balance the oxygen with H2O
• Balance the Hydrogen with H
• Balance the electrons by multiply the ½ rxns
• Add ½ and omit duplicates

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S2O62-(aq) HClO2(aq) ? SO42-(aq) Cl2(g)
Step 1 Write two unbalanced half-equations,
S2O62- ? SO42-
HClO2 ? Cl2
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Step 2 Insert coefficients to make the numbers of
atoms of all elements except oxygen and hydrogen
equal on the two sides of each half-equation.
S2O62- ? 2 SO42-
2 HClO2 ? Cl2
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Step 3 Balance oxygen by adding H2O to the side
deficient in O in each half-equation
S2O62- --gt 2 SO42-
2 H2O
(8Os)
(6Os)
(Need 2 Os)
2 HClO2 --gt Cl2
4 H2O
(4Os)
(Need 4 Os)
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Step 4 Balance hydrogen. For half-reaction in
acidic solution, add H on to the side deficient
in hydrogen.
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2H2O S2O62- ? 2SO42-
4H
(4 Hs)
Need 4 Hs
2HClO2 ? Cl2 4H2O
6H
Need 6 Hs
(2 Hs)
(8 Hs)
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Step 5 Balance charge by inserting e- (electrons)
as a reactant or product in each half-reaction.
Take into account the charge of the H
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2H2O S2O62- ? 2SO42- 4H 2 e-
Oxidation reaction, electrons are lost on the
product side
0
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6H 2HClO2 6 e- ? Cl2 4H2O
Reduction reaction, electrons are gained on the
reactant side
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Step 6 Multiply the two half-equations by numbers
chosen to make the number of electrons given off
by the oxidation equal to the number taken up by
the reduction.
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2H2O S2O62- ? 2SO42- 4H 2 e-
Oxidation reaction, electrons are lost on the
product side
0
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6H 2HClO2 6 e- ? Cl2 4H2O
Reduction reaction, electrons are gained on the
reactant side
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Step 6 Then add the two half-equations and cancel
out the electrons. If H ion, OH- ion, or H2O
appears on both sides of the final equation,
cancel out the duplication
6H2O 3S2O62- --gt 6SO42- 12H 6e-

6H 2HClO2 6e- --gt Cl2 4H2O
2H2O 3S2O62- 2HClO2 --gt 6SO42- 6H Cl2
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http//www2.hmc.edu/karukstis/chem21f2001/tutoria
ls/problemsRedoxFrame.html
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Balancing redox reaction handout. Handout 50
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These oxidation numbers are critical because they
let us know whether or not a electron is being
transferred and if the reaction is a redox
reaction and if we are creating an electric
current.
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We are talking about electricity

• So we have to talk about
• Luigi Galvani

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We know metals have an accumulation of electrons
and want to oxidize

• Or we can accumulate the electrons ourselves

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Laden jar
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One day during the 1800 some student discharged a
laden jar during a dissection

• Metal probe and the jump of a frog

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Galvani heard about it and postulated living
electricity
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Galvani ran the experiment
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Dissimilar metals
Volta disagreed
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Battery

• Copper with a stronger hold on electrons would
  actually drag electrons from the zinc.

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Copper will accumulate a negative charge and zinc
left with a positive charge
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Voltaic jump
Depending on the temperature more or less
electrons will move to the copper.
The influence temperature has on electron
transfer is the basis for the electric thermometer
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Show penny and nickel current
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When a stick of zinc (Zn) is inserted in a salt
solution, there is a tendency for Zn to loose
electron according to the reaction, Zn Zn2
2 e-.
Similarly, when a stick of copper (Cu) is
inserted in a copper salt solution, there is also
a tendency for Cu to loose electron according to
the reaction, Cu Cu2 2 e-.
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Remember,ions go into solution (they have a
charge so they are hydrated)nonionic substances
do not have a charge and are solids and can not
be hydrated
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Who is stronger at losing electrons,who clings
to more electron and what pressure is generated
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This electrical pressure is measured in volts

• Reduction (GER) occurs at the cathode
• Oxidation (LEO) occurs at the anode
• These reactions take place on the surface of the
  electrodes

Reduce the number of cats
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Charts have the electrical potential (volts) of
the ½ reactions so you know who wins
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The electrons cover the copper like ants on a
sugar cube and Zn2 goes into solution

• But this only lasts a second because quickly the
  Cu gets a charge Zn a and everything stops

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So they needed to figured a way to keep the Zn
solution neutral and Cu solution neutral so the
electron flow would keep going. English
chemist John Frederick Daniell developed a
voltaic cell in 1836 which used zinc and copper
and solutions of their ions.
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The electrolyte solution is composed of Zn2 Cu2
and SO4-2
As the CuSO4 acquires electrons it becomes Cu(s)
and the freed-up SO4-2 moves across the porous
barrier to neutralize the Zn2
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Electrolytic cell

• A simple combination of dissimilar metals that by
  the addition of a electrolyte are enabled to
  continuously transfer electrons from the
  oxidation side to the reduction side

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The transfer of electrons is the electric current
electrochemistry
Electrons leave the (they are crowded and move
to the (cathode) by way of an exterior wire
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http//hyperphysics.phy-astr.gsu.edu/hbase/chemica
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• The anode (oxidation) material, zinc (Zn) gives
  up two electrons (e-) per atom in a process
  called oxidation. This process leaves stable zinc
  ions (Zn2) behind. (An ion because of its charge
  is hydrated by the charged fluid (electrolyte).
  The anode has a lot of free electrons and is
  where the abundance of loose electrons cluster

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• AFTER the electrons move through the light bulb,
  they re-enter the battery at the cathode
  (reduction). There they combine with the
  cathode's active material, manganese dioxide
  (MnO2), in a process called reduction. MnO2 is a
  solid and acquires the electrons resulting in
  MnO2-.

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The presence of MnO2- at the cathode and the
presence of Zn2 at the anode splits the
neighboring water. H goes to the MnO2- and the
OH- goes to the Zn2. The water is important
because if the ions of MnO2 and Zn were not
neutralized then the reaction would stop as soon
as it started
The neutralizing ion in this case is water
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Remember electrons always want to shift to the
area of lower energy

• Electrical pressure

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Winter green lifesavers

• This phenomenon is the result of tiny electric
  sparks that occur when sucrose crystals in the
  Lifesaver crack as they are exposed to severe
  stresses. A separation of electric charge occurs
  between the two sides of the fracture tip and an
  electric discharge occurs through the air
  separating those two sides. The light that you
  see is produced by this electric discharge.
  Fractional luminicence

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 Rechargeable car battery

• This battery has two kinds of metal plates 1)
  lead 2) lead oxide but instead of salt water
  like volta used it has a strong acid called H2SO4

The H2SO4 being ionic dissociates in water into
2 H and SO4-
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Battery solution
This solution carries a current
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The lead attracted the negative SO4-2
Pb(s) SO4 -2 ? Pb SO4 2e-
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Creation of PbSO4 and release of 2 e
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At the PbO2 pole
PbO2(s) 4H(aq) SO4 -2 2e- ? Pb SO4 H2O
H attacks the O2 of the PbO2 creating H2O and
leaves Pb so any remaining SO4 attacks the Pb
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So the PbSO4 powder is deposited on both the
poles
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As the PbSO4 is deposited on both poles 2e move
from the Pb pole to the PbO2

• Pb(s) SO4-2 ? PbSO4 2e-
• PbO2(s) 4H(aq) SO4-2 2e-? Pb SO4 H2O
• Pb(s)PbO2(s)H2SO4(l)?PbSO4(s)H2OEnergy

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Pb(s) PbO2(s) H2SO4(l)?PbSO4(s) H2O energy

• If energy is added the equation is shifted to the
  left.

On the product (right side) of the equation the
H2O is cracked to H is released, the PbO2 is
reestablished and the H2SO4 is regained
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Standard Electrode PotentialsStandard Reduction
Potentials

• As we said earlier the pressure of the electrons
  is different for each oxidation or reduction
  reaction.

The standard used to describe a electrode
potential is the reaction for 2H3O(aq) 2e. ?
H2(g) 2H2O(l) Eo Standard electrode potential
.0000 volts
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Think of E as a measure of the ability of an
electrode to gain electrons ( reduction) .A more
positive value means the electrode is more likely
to be a cathode.
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Ecell Ecathode - Eanode

• The more the E is the cathode.

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Do the calculations for a number of Reduction
potentials
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electroplating

• The depositing of a metallic ion in water and the
  associated deposition of that ion on a negatively
  charged metal