Extra Credit 25

17.3.4

This reaction takes place at standard conditions: 25 °C and 1 M

Nitrate does not matter – it’s in both solutions

Step 1 – determine which one is getting oxidized and which one is getting reduced based on E values on the SRP chart (smaller E = oxidized=anode)

Cathode (Reduction) : \( Ag^+(aq) + e^- →Ag(s) \ E° = 0.7996V \)

Anode (Oxidation) : \(Zn^{2+}(aq) + 2e^-→Zn(s) \ E° = -0.7618 V \)

Oxidation reaction: \(Zn(s) → Zn^{2+}(aq) + 2e-\)

oxidation is loss of electrons so electrons should end of on the product side

Step 2 – For the overall reaction, combine the equations for the Cathode and Anode

multiply the reduction equation by 2 so the electrons will cancel out

\(2Ag^+(aq) + Zn(s) → 2Ag(s) + Zn^{2+}(aq)\)

Step 3 – calculate final V cell

\(E \ Cathode: \ .7996 V \ E \ Anode: \ -.7618 V\)

\(E \ cell = E \ cathode (reduction) – E \ anode(oxidation)\)

\(E \ cell = 0.7996 -(- 0.7618) = 1.5614 V \)

E cell = 1.5614 V (We know that it is a spontaneous reaction because of the positive E cell)

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19.1.23

Initially as cyanide ions are added, the first reaction takes place.

Ag⁺(aq)+CN^-(aq)⟶AgCN(s)

The sodium and nitrate can be ignored since they are spectator ions, meaning they do not participate in the reaction.

As more CN is added, it binds to the AgCN -> which is a soluble compound

AgCN(s)+CN^-(aq)⟶[Ag(CN)2]^-(aq)

The net reaction is \(Ag^+(aq)+2CN^-(aq)⟶[Ag(CN)_2]^-(aq)\)
The CN^- ions then attach to the solid AgCN to create Ag(CN)2)-. The product is, Ag(CN)2 – which is aqueous, explaining why the precipitate disappears.

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12.4.16

a. Important info: Fluorine 18, half-life = 109.7 min

Use formula to find rate constant \( Kt= - ln (.5) \)

Convert minutes to seconds: \(109.7 min * 60 sec/min = 6582\ sec\)

Plug in and solve: \( K = - ln (.5)/t \)

\( K = - ln (.5)/6582 \)

\( K = 1.053*10^{-4} \ sec^{-1}\)

b. Important info: \( K = 1.053*10^{-4} \ sec^{-1} \) from part a, t = 5.59 hours, amount left = ?

Use formula to find final amount: \( Kt_x= - ln ([A]/[A_0\)])

Convert hours to seconds: \(5.59 \ hours * 60\min/hour * 60 \ sec/min = 20124 \ sec)

Plug in and solve: \( Kt_x= - ln ([A]/[A_0\)])

\( 1.053*10^{-4}*20124= - ln ([A]/[A_0\)])

\(e^{-2.119}\) = [A]/[A₀] = \( .1201)\)

([A]/[A₀]) = 12.01%

c. Important info: \( K = 1.053*10{-4} \ sec^{-1} \) from part a, looking for time for 99.99% to decay

Use formula to find time: \( Kt_x= - ln ([A]/[A_0\)])

Amount remaining = .01% = .0001, initial = 1

Plug in and solve: \( t_x= - ln ([A]/[A_0]/K\)

\( t_x= - ln (.0001/1)/1.053*10^{-4} \)

Time = \(87467.61 sec * frac(1 min / 60 sec) * frac(1 hour / 60 min) = 24.29 \ hours \)

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21.2.10

Step 1 – find the number of protons and neutrons in each isotope

Step 2 – look for isotopes that have the magic number of protons OR neutrons– these are extra stable – if they have magic numbers of protons AND neutrons, they are even more stable. The magic numbers are: 2, 8, 20, 50 , 82, and 126

If you cannot decide stability yet, continue to the next step.

Step 3 – Ratio Test (check the ratio between Neutrons and protons)

For elements 1-20 if this number is 1, then it is likely to be stable. This means that for elements from 1-20 if the ratio is not near 1, the element is not stable.

For elements 11 – 20 – the ratio test is not always correct.

For elements beyond this range, the ratio increases to around 1.5. This means that for elements 21-83 if the ratio is greater than 1.5, the element is not stable.

Elements beyond element 83 are rarely stable.

If you cannot decide stability yet, continue to the next step.

Step 4 – If the number of protons and the number of neutrons are both even, likely to be stable.

If the number of protons and the number of neutrons is odd, it is likely to be unstable. (ratio rule is more important than this, so if it meets the ratio rule it is stable)

Step 5 – if you have the chart on hand, cross check to confirm your results. The chart is found here.

1. argon-40

Step 1: 18 Protons, 22 Neutrons

Step 2: no magic numbers

Step 3: Ratio = 22/18 = 1.22 close to 1

Step 4: Both Even

Stable

Step 5: Cross check (confirmed)

2. oxygen-16

Step 1: 8 Protons, 8 Neutrons

Step 2: 8 is a magic number

Stable

Step 5: Cross Check - confirmed

3. \(^{122}Ba \)

Step 1: 56 Protons, 66 Neutrons

Step 2: no magic numbers

Step 3: Ratio = 66/56 = 1.17 -> not close to 1.5

Unstable

Step 5: Cross Check –confirmed

d. \(^{58}Ni \)

Step 1: 28 Protons, 30 Neutrons

Step 2: no magic numbers

Step 3: Ratio = 30/28 = 1.07

Step 4: Both even numbers

Stable

Step 5: Cross Check – confirmed

5. \(^{205}Tl \)

Step 1: 81 Protons, 124 Neutrons

Step 2: no magic numbers

Step 3: Ratio = 124/81 = 1.53 close enough to 1.5

Stable

Step 5: Cross Check – confirmed

6. \(^{210}Tl \)

Step 1: 81 Protons, 129 Neutrons

Step 2: no magic numbers

Step 3: Ratio = 129/81 = 1.59 > 1.5 (close but inconclusive)

Step 4: Both odd numbers

Unstable

Step 5: Cross Check – confirmed

7. \(^{226}Ra \)

Step 1: 86 Protons, 140 Neutrons

Step 2: no magic numbers

Step 3: Ratio does not matter, 86 > 83

Unstable

Step 5: Cross Check – confirmed

8. magnesium-24

Step 1: 12 Protons, 12 Neutrons

Step 2: no magic numbers

Step 3: Ratio = 12/12 = 1

Stable

Step 5: Cross Check – confirmed

Isotopes a, b, d, e, h lie within the band of stability.

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21.7.2

Radon's primary decay is alpha decay. People's skin protect them from alpha radiation, but when inhaled, the alpha decay damages the tissue in the nose, throat, and lungs which can cause severe damage, cancer, and possible result in death.

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20.4.12

In a cell diagram | is used to separate the states of matter within a cell, and || is used to separate the 2 cells

SHE = Standard Hydrogen Electrode \(H^2 \ (g) \) and \(Pt \ (s) \)

Step 1 - look at the reaction, identify which elements are oxidized and reduced

standard cell diagrams are written as\( oxidation \ cell \ (anode) \ || reduction \ cell \ (cathode) \)

\(Zn(s) → Zn^{2+} + 2e^-\) - oxidation reaction (electrons are on the right side of the equation)

\(2H^+ (aq) + 2e^1 → H_2(g) \) - reduction reaction (electrons are on the left side of the equation)

\(Zn(s)|Zn^{2+}(aq)(1M)||H^+ (aq)(1M)|H_2(g)(1atm)|Pt(s)\)

- assume 1M and 1atm b/c std. conditions

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20.4.14

a. \(Cu^+(aq) + Ag^+(aq) → Cu^{2+}(aq) + Ag(s)\)

Step 1: Check if it’s balanced (both sides of equation have the same # of each element and same overall charge)

1 Cu on each side, 1 Ag on each side, overall charge +2 on each side: BALANCED SHE; so there is no need to balance the equation using the half-reaction method.

Step 2: Determine the oxidation and reduction reactions

Oxidation: \( Cu^+(aq)→ Cu^{2+}(aq) + e^{-} \) (oxidation is loss of electrons)

Reduction: \(Ag^+(aq) + e^{-} →Ag(s)\) (reduction is gain of electrons)

Step 3: To find the standard reduction potential use E cell = E cathode - E anode

\(E \ cathode = E \ reduction = 0.7996\)

\(E \ anode = E \ oxidation = 0.159\)

Plug these values in the formula: \(0.7996-0.159 = 0.6406 V\)

b. \(Sn(s) + Fe^{3+}(aq) → Sn^{2+}(aq) + Fe^{2+}(aq)\)

Step 1: Check if it’s balanced (both sides of equation have the same # of each element and same overall charge)

1 Sn (yes), 1 Fe (yes), overall charge +3 on left +4 on right: NOT BALANCED

Step 2: Separate into half reactions, decide which one is oxidation and which one is reduction, and balance the reaction

Oxidation: \( Sn(s)→ Sn^{2+}(aq)+2e^- \) (oxidation is loss of electrons)

Reduction: \(e^- + Fe^{3+}(aq) →Fe^{2+}(aq)\) (reduction is gain of electrons)

Balance: multiply the reduction reaction by 2, so the e- will cancel when the equations are added

Net Reaction: \(Sn(s) + 2Fe^{3+}(aq) →2Fe^{2+}(aq) Sn^{2+}(aq)\)

Step 3: To find the standard reduction potential use /(E cell = E cathode - E anode/)

\(E \ cathode = E \ reduction = 0.771\)

\(E \ anode = E \ oxidation = -0.14\)

Plug these values in the formula: \(0.771-(-0.14) = 0.911 V\)

c. \(Mg(s) + Br_2(l) → 2Br^−(aq) + Mg^{2+}(aq)\)

Step 1 - Check if it’s balanced (both sides of equation have the same # of each element and same overall charge)

1 Mg on both sides, 2 Br on both sides, net charge 0 on both sides: BALANCED

Step 2 - Identify oxidation and reduction reactions

Oxidation: \( Mg(s)→ Mg^{2+}(aq) + 2e^{-} \) (oxidation is loss of electrons)

Reduction: \( Br_2(l) + 2e^{-} → 2Br^−(aq)\) (reduction is gain of electrons)

Step 3: To find the standard reduction potential use /(E cell = E cathode - E anode/)

\(E \ cathode = E \ reduction = 1.087\)

\(E \ anode = E \ oxidation = -2.356\)

Plug these values in the formula: \(1.087-(-2.356) = 3.443 V\)

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14.1.3

General Trends:

Increase in concentration → increase in reaction rate (As concentration increases the reactants have a higher probability of colliding with another reactant particle, inducing a chemical reaction.)

Increase of temperature → increase in reaction rate (As temperature increases, this makes the particles move at a faster speed so when the particles collide, they collide with more energy and are more likely to reach the activation energy threshold)

Presence of a catalyst → increase in reaction rate (A catalyst by definition is a substance that decreases the activation energy, without being used up in the reaction)

Physical properties of the reactants → increase in reaction rate (If the reactants have the same physical properties, they are more likely to react)

Physical and chemical properties of the solvent: depends on the property

High solubility → increase in reaction rate

Solvent polarity → depends

Increasing solvent viscosity → decrease in reaction rate