Chem exam 3 chapter 20

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Last updated 7:54 AM on 5/6/26
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Radioactive decay types

  1. If on Right and it’s -1 for Z (atomic charge) it’s….

  2. If on left and it’s -1 for Z, it’s…

  3. If it’s -4 and -2, the element identity on right is Helium, and it’s…

  4. If it’s +1 to Z on right, it’s…

  1. On right, -1 = positrion emission

  2. On left, -1 = Electron capture

  3. -4 and -2 (helium) = alpha particle decay

  4. +1 on right to Z, is beta particle decay

Review 20.1 if you don’t get it but it’s pretty simple once you get hang of it.

Top number is molar mass

Bottom number if atomic number from periodic table

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ID unstable Isotopes and predict their modes of decay for

37Cl, 222Rn, 26Mg, 66Co, 116I

  1. ID # of protons and neutrons in each element

a. Proton = atomic #

b. Neutron = The number above the element - protons

  1. ID stability

  • More than 82 protons = non-stable

  • Odd # of Protons/Neutrons = non-stable

  • Even number of Protons/Neutrons = Stable

  • Magic # for either proton or neutron = Stable

Protons: 2, 8, 20, 28, 50, or 82

Neutrons: 2, 8, 20, 28, 50, 82, or 116

  1. For those that are unstable, predict the mode of decay

  • If protons are greater than 82, it’s alpha decay.

  • Compare value above the unstable isotope to its atomic mass on periodic table. If the value above the unstable isotope is higher than the number for its atomic mass, it will be beta decay. (above belt)

  • Compare value above unstable isotope to its atomic mass, if the value above the unstable isotope is lower than its atomic mass in periodic table, you can predict it will be positron emission or electron capture (below belt)

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Beta particles have a mass number of ___.

Beta particle formation is accompanied by the conversion of a _____ into a _____.

Beta particles have mass numbe of 0

Beta particle formation is accompanied by the conversion of a neutron into a proton.

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List position and values for Alpha, Beta, Positrion Emission, and Electron capture

Alpha: Right, subtract 4 and 2 for He
Beta: Right, add 1 to Z
Positrion emission: Right, subtract 1 from Z
Electron capture: Left, subtract 1 from Z

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The mass of a nucleus is always _____ the sum of the masses of its nucleons

mass of nucleus is always less than sum of its nucleons

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What’s the equation for finding half lives


ln(initial/final) = (ln2/half life)t

If you gotta solve for half life, what do you gotta do

mutliply 2 by ln2, then solve for half life

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Of elements that have non-radioactive isotopes and are stable, which has highest atomic #?
- lead

  • iron

  • - radium

  • uranium

Answer is lead

It’s basically asking, which is going to be stable.

Radium and uranium have more than 82 protons based on atomic number, therefore they’re not stable.

Lead has highest atomic number compared to iron, and is still stable, therefore its lead.

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As soon as you see Voltaic cell, what does this tell you?

Spontaneous redox reaction

+E and negative G

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Voltaic/Galvanic (same thing) cells must be ______.
The one that’s Oxidized will have more ____ E value

The one that’s Reduced (cathode) will have a more ____ E-cell

Voltaic/Galvanic cells MUST be Spontaneous! (+E, -G, K>1)

Oxidized will have more negative E value in Voltaic cells

Reduced (cathode) will have a more positive E value in voltaic cells

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Write overall reaction between AgNO3 and Solid Zn in a Voltaic cell

  1. Identify the Anode and Cathode

  • Anode = more negative E value in Voltaic/Galvanic cell

  • Cathode = more positive E value in Voltaic/Galvanic cell

Zinc —> Zinc(s)

AgNO3 —> Ag+ + NO3-

  1. Write out Anode

Zn(s) —> Zn+2 + 2e-

Write out Cathode

Ag+ + e- —> Ag(s)

Balance out electrons

2Ag + 2e- —> 2Ag(s)

Assemble net ionic equation

Zn + 2Ag —> Zn+2 + 2Ag

Add back in the Spectator ions

Zn(s) + 2AgNO3 —> Zn(NO3)2 + 2Ag

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<ol><li><p>ID the Anode and Cathode for the voltaic cell</p></li><li><p>Balance out the electrons for both, multiply if necessary to get them to cancel out</p></li><li><p>Write out the net ionic equation where the charges are equal (i.e +2 = +2)</p></li><li><p>Write the overall equation where you plug back in the placeholders on the values that have charges to get them to cancel out</p></li></ol><p>i.e<br><br>from the start<br>Mg —&gt; Mg+2 + 2e-<br>Cu+2+ 2e- —&gt; Cu(s)<br><br>Net ionic equation</p><p>Mg + Cu+2 —&gt; Mg+2 + Cu(s)<br><br>we have a +2 on Cu, and a +2 on Mg</p><p>and we want overall charge on both sides to be 0</p><p>So we add back in the placeholder value of NO3- onto both, but NO3- is only -1, so we need 2. so it becomes</p><p></p><p>Mg(s) + Cu(NO3)<sub>2</sub> —&gt; Mg(NO3)<sub>2</sub> + Cu(s)</p><p></p>
  1. ID the Anode and Cathode for the voltaic cell

  2. Balance out the electrons for both, multiply if necessary to get them to cancel out

  3. Write out the net ionic equation where the charges are equal (i.e +2 = +2)

  4. Write the overall equation where you plug back in the placeholders on the values that have charges to get them to cancel out

i.e

from the start
Mg —> Mg+2 + 2e-
Cu+2+ 2e- —> Cu(s)

Net ionic equation

Mg + Cu+2 —> Mg+2 + Cu(s)

we have a +2 on Cu, and a +2 on Mg

and we want overall charge on both sides to be 0

So we add back in the placeholder value of NO3- onto both, but NO3- is only -1, so we need 2. so it becomes

Mg(s) + Cu(NO3)2 —> Mg(NO3)2 + Cu(s)

s

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How do you know which aqueous ion concentration to increase to drive the reaction forward?

Take the Magnesium/Cadmium reaction you just solved:

Mg(s)+Cd2+(aq)Mg2+(aq)+Cd(s)Mg(s) + Cd^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cd(s)

Driving the reactant ion conc forward will increase forward direction for more voltage, more E, more negative G, more K!

vice versa for product ion conc

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<p>For Electrolysis, you have 2 reactions fighting for cathode, and 2 fighting for anode</p><p></p><p></p>

For Electrolysis, you have 2 reactions fighting for cathode, and 2 fighting for anode

I.E here, it’s Water reaction vs Br reaction

and Water reaction vs K reaction

  1. ID which is the Anode and Cathode by lookiong at the sign of the element

I.E KBr in H2O

K = +, therefore If we write it as K+ —> K(s) it would be reduced

Br = -. therefore if we write it as Br- —> Br, it would be oxidized

  1. Look at the 2 reactions that are displaying Oxidation , choose the one that is more negative.

This is the anode!

  1. Look at the 2 reactions that are displaying reduction, choose the one that is more positive

That will be your

Cathode!

So ID which are anode and cathode by looking at question and assigning signs to the molecular formula i.e KBr = K+ --> K(s) reduction (cathode) and Br- --> Br2 (anode) oxidation

2. For choosing oxidation vs whatevers in solution like water, choose the one that's more negative. I.E, if Br is -0.5, and H2O is -0.6, then H2O will be anode

3. For choosing reduction vs whatevers in solution like water, choose one thats more positive. If K is +47, and h2o is + 2, choose K as cathode.

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Which pair of substances will produce the greatest EMF when used as electrodes in 1M solutions of their ions

Reactivity

Na > Zn > Fe > Pb > H > Cu > Hg> Ag

The most reactive will be the anode, and have the most negative reductipn potential

The least reactive will be further out to left and have most positive reduction potential

i.e Na + Ag would have greatest reactivity together

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Most reactive = ______, and has most _____ reduction potential

Least reactive = _____, and has most _____ reduction potential.


Most reactive = Anode, and has most negative reduction potential
Least reactive = Cathode, and has most positive reduction potential

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Strongest Oxidizng agent =

Strongest Reducing agent =

What about on Standard Oxidation potentials chart?

Strongest oxidizng agent = Most positve value

Strongest reducing agent = Most negative value

ON STANDARD REDUCTION POTENTIALS CHART

ON STANDARD OXIDATION POTENTIALS CHART

Strongest Oxidizing agent = negative

Strongest reducing agent = positive

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cancel log

cancel ln

cancel log

10value

cancel ln

evalue

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Equation for amps and mass and shit

m = I x t x MM / nF

m = mass

i = amps

t = time

MM = molar mass

n = electrons

F = 96485

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For a Concentration cell, what is the value of Ecell always???

Why is this important for determining cell potential?

Ecell is 0.

Still need to use nernst equation to solve, but you can remove the Ecell from the equation.

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  1. Write out equation from anode/cathode

  2. ID the reactant ion

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