Copper Cycle Lab Report
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The Copper Cycle
Samantha Delaunay
Cory Silguero
The University of Texas at Austin
1
P. Sotelo SP2024
Redox reactions stand as prominent methods for producing valuable compounds. In light of increasing awareness of green chemistry, there's been a search for sustainable routes to redox reactions. This has led to significant interest in photoredox catalysis, a burgeoning field where molecules undergo oxidation or reduction using visible light – a vast natural resource. This tutorial paper offers an introductory overview of photoredox catalysis, accompanied by select examples, with the aim of inspiring further exploration and advancement in this rapidly evolving research domain.
In a comprehensive study, researchers investigated the oxygen redox reaction and structural reversibility of NCM or NCA, promising cathode materials for high-energy-density Li-
ion batteries. Utilizing synchrotron X-ray absorption spectroscopy, scanning transmission electron microscopy, and density functional theory calculations, they uncovered that oxygen redox occurs due to cation mixing upon delithiation, despite the absence of Li-O-Li configurations in its pristine form. The formation of the I41 structure was linked to the migration
of transition metals in deeply delithiated material, extending the phase transformation route from layered to rock-salt structure. These findings deepen the understanding of oxygen redox and its impact on structural transformations in Ni-rich layered oxides, offering insights for enhancing cathode material performance in next-generation Li-ion batteries.
Antoine Lavoisier's dedication to precise measurement was evident in his meticulous experiments. In one experiment, burned phosphorus and observed the formation of a white flaky product. Although the phosphorus, the reactant in this case, was not soluble in water, the product was. Therefore, he collected the product, separating it from the unreacted phosphorus through thorough washing with water. After drying, he could measure the amount of phosphorus that had
burned, the quantity of oxygen consumed, and the mass of the resulting product. Through his experiments, he consistently found that the mass of the product equaled the sum of the masses of the consumed reactants. This discovery laid the foundation for the law of conservation of mass.
Examining the Hindenburg disaster of 1937 provides a chemical perspective on oxidation-reduction reactions. The dirigible, filled with hydrogen, utilized this gas to maintain buoyancy. Unfortunately, during its voyage from Germany to the United States, the airship met a
tragic end due to the explosive reaction between hydrogen and oxygen. In this event, hydrogen lost electrons, undergoing oxidation, while oxygen gained electrons, experiencing reduction.
Metallic copper will undergo a sequence of reactions with different reagents to generate four distinct copper compounds, followed by its restoration to its original metallic state. Each reaction's chemical and physical alterations will be observed and documented, with balanced chemical equations being formulated and categorized. The final step involves determining the percent recovery of the initial copper product.
2
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Table 1. Initial and final masses of Cu
(s). Description
Mass (g)
Cu
(s)
wire
0.361
Evaporating dish
25.906
Evap. dish + Cu
(s)
product
26.193
Cu
(s)
product
0.287
Table 2.
Observations for Reactions 1-5.
Reaction
Product
Observations
1
Cu(NO
3
)
2
Emits brown gas, turns a blue color 2
Cu(OH)
2
Neon/bright blue, pH paper is blue on first test
3
CuO
Bright blue to dark blue to black cloud
4
CuSO
4
Light blue, burnt Cheeto wire
5
Cu
(s)
Orange copper precipitate, stinky
3
P. Sotelo SP2024
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For the first step of the copper cycle, copper metal (Cu) reacts with nitric acid (HNO3) to form copper(II) nitrate (Cu(NO3)2). The initial reddish-brown metallic copper turns into a blue solution of copper(II) nitrate. Figure 1:
End of Step 1 Copper Cycle
Next, Copper(II) nitrate (Cu(NO3)2) reacts with sodium hydroxide (NaOH) to form copper(II) hydroxide (Cu(OH)2) A blue solution of copper(II) nitrate reacts with colorless sodium hydroxide to form a blue-green precipitate of copper(II) hydroxide. Figure 2:
End of Step 2 Copper Cycle
4
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Thirdly, Copper(II) hydroxide (Cu(OH)2) decomposes upon heating to form copper(II) oxide (CuO). The blue-green liquid of copper(II) hydroxide turns into a black solid of copper(II) oxide with a clear liquid. Figure 3:
End of Step 3 Copper Cycle
After this, Copper(II) oxide (CuO) reacts with sulfuric acid (H2SO4) to form copper(II) sulfate (CuSO4) and water (H2O). The black solid of copper(II) oxide dissolves in the sulfuric acid solution to form a blue solution of copper(II) sulfate. Figure 4:
End of Step 4 Copper Cycle
5
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Finally, Copper(II) sulfate (CuSO4) reacts with metallic aluminum (Al) to form metallic copper (Cu). The blue solution of copper(II) sulfate reacts with metallic aluminum to deposit metallic copper onto the aluminum surface and the liquid will become more clear as the copper forms. Figure 5:
End of Step 5 Copper Cycle
The formation of precipitates in some reactions is attributed to the low solubility of certain copper compounds. Insoluble or sparingly soluble compounds tend to precipitate out of solution, whereas soluble compounds remain dissolved.
Step 1 is a redox reaction, step 2 is a precipitation reaction, step three is a decomposition reaction, step 4 is an acid-base reaction, and step 5 is also a redox reaction. Their according chemical equation requirements are described below.
Step 2 Precipitation Reaction:
Balanced Formula Unit:
Cu(NO
3
)
2 (aq) + 2 NaOH
(aq) → Cu(OH)
2 (s) + 2 NaNO
3 (aq)
Total Ionic:
Cu
+2
(aq) + 2 NO
3
-
(aq) + 2 Na
+
(aq) + 2 OH
-
(aq) → Cu(OH)
2 (s) + 2 Na
+
(aq) + 2 NO
3
-
(aq)
Net Ionic:
Cu
2+
(aq) + 2 OH
-
(aq)
→ Cu(OH)
2 (s)
Step 4 Acid-Base Reaction:
Balanced Formula Unit:
CuO
(s)
+ H
2
SO
4 (aq) → CuSO
4 (aq) + H
2
O
(l)
6
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Total Ionic:
CuO
(s)
+ 2 H
+
(aq) + SO
2-
4 (aq) → Cu
2+ (aq) + SO
2-
4 (aq) + H
2
O
(l)
Net Ionic:
CuO
(s)
+ 2 H
+
(aq) → Cu
2+ (aq) + H
2
O
(l)
Step 3 Decomposition Reaction:
Balanced Chemical Equation:
Cu(OH)
2 (s)
→ CuO
(s) + H
2
O
(1)
Step 1 Redox Reaction:
Balanced Chemical Equation:
Cu
(s)
+ 4 HNO
3 (aq) → Cu(NO
3
)
2 (aq) + 2 NO
2 (g) + 2 H
2
O
(l)
Oxidizing Half Reaction: Cu
(s) → Cu
2+
(aq) + 2e
-
Reduction Half Reaction: 4 H
+
(aq) + 4e
- + 2 NO
3
-
(aq) → 2 NO
2 (g) + 2 H
2
O
(l)
Oxidizing Agent: Nitric Acid (HNO
3
)
Reduction Agent: Copper (Cu)
Step 5 Redox Reaction:
Balanced Chemical Equation:
3 CuSO
4 (aq)
+ 2 Al
(s) → 3 Cu
(s) + Al
2
(SO
4
)
3 (aq)
Oxidizing Half Reaction: 2 Al
(s) → 2 Al
3+
(aq) + 6e
-
Reduction Half Reaction: 3 Cu
2+ (aq) + 6e
- → 3 Cu
(s) Oxidizing Agent: Copper Sulfate (CuSO
4
)
Reduction Agent: Aluminum (Al)
Based on the balanced chemical equation for Reaction 5, the compounds being removed are copper sulfate (CuSO
4
) and aluminum sulfate (Al
2
(SO
4
)
3
). Ethanol has a relatively low boiling point and evaporates quickly at room temperature. By rinsing the copper product with ethanol, residual water from previous washing steps can evaporate more effectively, aiding in drying. Ethanol is generally considered compatible with copper and does not react chemically with it under normal conditions. This ensures that the washing solvent does not introduce any additional contaminants or react with the copper product.
7
P. Sotelo SP2024
A steam bath was the chosen method to dry the copper product because it prevents the oxidation of the copper product by avoiding direct exposure to atmospheric oxygen.
The percent recovery of the copper is 79.5%. The Law of Conservation of Mass is a fundamental principle in chemistry that states that in a chemical reaction, mass is neither created nor destroyed. The total mass of the reactants before a chemical reaction is equal to the total mass of products after the reaction. If the percent recovery is not 100%, it may be caused by an incomplete drying process. If residual water is left behind,
it can lead to an overestimation of its mass during the weighing process. Another reason is the overheating of copper during the drying procedure. This can lead to the oxidation of the product through atmospheric oxygen. Percent Recovery:
(0.287 g / 0.361 g) x 100% = 79.5%
If the percent recovery is above 100%, it may be caused by the contamination of impurities in the final copper recovered which can lead to the overestimation of the final mass.
Another reason is the loss of copper product during the various chemical reactions along the way. If the percent recovery is below 50%, it may be caused by an early completion of step 5. If the copper being recovered on the aluminum wire has not fully formed and was taken out too early, it will be a lower mass than the original copper wire. Another reason is the loss of product during the washing process. 8
P. Sotelo SP2024
REFERENCES 1.
Cauwenbergh, Robin, and Shoubhik Das. “Photocatalysis: A Green Tool for Redox Reactions.”
Synlett
, Georg Thieme Verlag KG, 9 June 2022, www.thieme-connect.com/products/ejournals/abstract/10.1055/s-0040-1706042
.
2.
Li, Shuwei, et al. “Anionic Redox Reaction and Structural Evolution of NI-Rich Layered Oxide Cathode Material.”
Nano Energy
, Elsevier, 4 May 2022, www.sciencedirect.com/science/article/abs/pii/S221128552200413X?via%3Dihub
.
3.
“Lavoisier.”
Chemistry LibreTexts
, Libretexts, 13 June 2023, chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(
Eames)/Chemistry_Basics/Lavoisier.
4.
Web Solutions LLC. “Oxidation-Reduction Reaction - History, Oxidation Numbers, Corrosion, Biological Processes, Current and Future Uses - Examples of Oxidation-
Reduction Reactions.”
History, Oxidation Numbers, Corrosion, Biological Processes, Current And Future Uses - Examples of Oxidation-Reduction Reactions - JRank Articles
, 2023, science.jrank.org/pages/4964/Oxidation-Reduction-Reaction.html.
9
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o 5 10 15 20 25
[Source: © International Baccalaureate Organization 2019]
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The experimental rate law is rate =
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Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine
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(2) 203(g) → 30,(g)
(3) 0,(g) + Cl(g) → CIO(g) + 0,(g)
AHn = ?
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(1) C10(g) + 0,(g) -
(2) 20;(g)
(3) 0,(g) + Cl(g) → ClO(g) + 0, (g)
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-122.8 kJ
>
=
→ 30,(g)
AHixn
= -285.3 kJ
→ ClO(g) + 0,(g)
AHixn
= ?
kJ
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-1000
Decomposition Rezctions
Carbon Reduction Reactions
AH
AS
(kJ)
(JK-mol)
O 2 Ag,0(s) → 4 Ag(s) + O2(g)
+62.2
+132.7
O 2 MgO(s) → 2 Mg(s) + O2(g)
+1202.4
+216.7
1000-
O 2 AlhO3(s) → 4 Al(s) – 3 02(g)
+3351.4
+546.8
O 2 Fe2O3(s) → 4 Fe(s) – 3 O2(g)
+1648 4
+5515
O 2 HgO(s) → 2 Hg(t) + 0,(g)
+181.7
+286.8
2000-
O 2 Cr,03(s) → 4 Cr(s)– 3 0,(g)
2269 4
+548.4
Temperature
3000
300
300 K
2000
Temperature ((K)
Clear Graph
The metals Mg and Al, like most metals in groups 1A, 2A, and 3A of the periodic table, cannot be obtained by chemical reduction with carbon at reasonable
temperatures. What procesS Is used to obtain these metals from their ores?
positive
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- Thank you in advance for your help!arrow_forwardЗа. Bromine and methanoic acid react in aqueous solution. Br, (aq) + HCO0Н (ад) — 2Br (aq) + 2H" (аq) + Со, (g) The reaction was monitored by measuring the volume of carbon dioxide produced as time progressed. o 5 10 15 20 25 [Source: © International Baccalaureate Organization 2019] 30 20 40 60 80 100 120 140 160 Time / s [Source: © International Baccalaureate Organization 2019] Determine from the graph the rate of reaction at 20 s, in cm³ s1, showing your working. 3b. Outline, with a reason, another property that could be monitored to measure the rate of this reaction. 3c. Describe one systematic error associated with the use of the gas syringe, and how the error affects the calculated rate. 3d. Identify one error associated with the use of an accurate stopwatch. Volume of carbon dioxide / cm T m mm moarrow_forwardWhat is the Qc value for this equation? Which direction will the reaction shift? Given: CO(g)+H2O(g)=CO2(g)+H2(g) Kc = 1.0 • [CO2(g)]= 2.0M • [H2(g)]= 2.0 M • [CO(g)]= 1.0 M • [H2O(g)]= 1.0 Marrow_forward
- 8 Reactions between certain organic (alkyl) halides and water produce alcohols. Consider the overall reaction for t-butyl bromide (2-bromo-2-methylpropane). (CH3)3CBr(aq) + H2O(1) →(CH3)3COH(aq) + H*(aq) + Br "(ag) k[(CH3)3CB1]. The accepted mechanism for the The experimental rate law is rate = %3D reaction is shown below. (1) (CH3)3C-Br(aq) →(CH3)3C*(aq) + Br "(aq) [slow] (2) (CH3)3C*(aq) + H2O(1) →(CH3)3C-OH2*(aq) [fast] (3) (CH3)3C-OH2*(aq) →H*(aq) + (CH3)3C-OH(aq) [fast] a) Why doesn't H2O appear in the rate law? b) Write rate laws for the elementary steps. c) What intermediates appear in the mechanism? d) Show that the mechanism is consistent with the experimental rate law. ed to the of consumption of Narrow_forwardThe reaction between peroxodisulphate (VI) ions, S2O8^2- and iodide ions, I- can be catalyzed by iron(III) ions, Fe^3+. a). Suggest a mechanism for the catalytic reaction b). Sketch energy profile diagram for the catalyzed and uncatalyzed reaction s..ns.arrow_forwardplease help and make things clear thank you so much (:arrow_forward
- An unknown inorganic phosphate undergoes first-order dehydration in a drying oven. Initially, the excess water content is measured to be 2.20 mg. After 2.30 days the water content is reduced to 1.24 mg. After 4.30 days, water content is only 0.751 mg. Determine the half-life of hydration under these conditions. Give the answer to 3 significant figuresarrow_forwardConsider the reaction a A(aq) → bB(aq) (1) Where a and b are stoichiometric coefficients for the balanced chemical reaction The average reaction rate r can be measured if the concentration change of the reagents or products is known: 1 ΔΙΑ 1 Δ r = and r = At b At a where A and B notate the concentration of A and B in solution, and A is always defined as the difference between "final" and "initial": A[4] = [A],2 - [4],1 and At = t2 – t . - A particular case of eq. (1) could be A(aq) → 2 B(aq) If the initial concentration of B is 0.2569 mol L1, and after 415.6 s it increases to 0.4633 mol L1, what is the rate of the reaction in mol L1s? mol L-1 s1. (Use scientific notation)arrow_forwardThe enzyme urease increases the rate of urea hydrolysis at pH 8.0 and 20 °C by a factor of 1014. Suppose that a given quantity of urease can completely hydrolyze a given quantity of urea in 19 minutes at pH 8.0 and 20 °C. How long would it take for this amount of urea to be hydrolyzed in the absence of urease at the same temperature and pH in sterile conditions? Include two significant figures in your answer. timeuncatalyzed years * TOOLS x10arrow_forward
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