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Copper forms two different cations, cuprous (Cu+) and cupric (Cu2+). When solid copper reacts with silver nitrate solution, two reactions are possible, as shown in equations a and b. Molecular Equation: Cu(s) + AgNO3(aq) → CuNO3(aq) + Ag(s) Net Ionic Equation: Cu(s) + Ag+(aq) → Cu+(aq) + Ag(s) Molecular Equation: Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s) Net Ionic Equation: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) 1. What mass of metallic silver can form from 1.147 g of copper metal according to Equation a? 2. What mass of metallic silver can form from 1.147 g of copper metal according to Equation b? 3. A student reacts 1.147 g of copper with a solution containing excess silver nitrate and recovers 1.914 g of silver metal. Which reaction, a) or b), most likely occurred in the experiment? Explain your answer completely. 4. When copper metal and silver ion react, which reactant is oxidized and which is reduced? 5. When copper metal and silver ion react, which is the more active metal?

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These are five questions and five complete answers.

Question 1. What mass of metallic silver can form from 1.147 g of copper metal according to Equation a?

Answer: 1.352 g Ag

Explanation:

1. Molecular equation and net ionic equations: given

  • Molecular: Cu(s) + AgNO₃(aq) → CuNO₃(aq) + Ag(s)
  • Net Ionic: Cu(s) + Ag⁺(aq) → Cu⁺(aq) + Ag(s)

2. Mole ratios: 1 mol Cu(s) : 1 mol AgNO₃(aq) : 1 mol Ag(s)

3. Convert 1.147 g of copper metal to number of moles:

  • n = mass in grams / atomic mass
  • n = 1.147 g / 63.546 g/mol = 0.01805 mol Cu (keep four signiticant digits such as the mass in grams is measured)

4. Use proportionality

  • 1 mol Cu / 1 mol Ag = 0.01805 mol Cu / x

       ⇒ x = 0.01805 mol Cu × 1 mol Ag / 1 mol Cu = 0.01805 mol Ag.

5. Convert 0.01805 mol Ag to grams

  • mass in grams = number of moles × atomic mass of Ag
  • mass in grams = 0.01805 mols × 74.922 g/ mol = 1.352 g Ag (keep 4 significant digits)

Answer: 1.352 g Ag

Question 2. What mass of metallic silver can form from 1.147 g of copper metal according to Equation b?

Answer: 2.705 g Ag(s)

Explanation:

1. Molecular and net ionic equations (given):

  • Molecular: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
  • Net Ionic: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

2. Mole ratios: 1mol Cu(s) : 2 mol AgNO₃(aq) : 2 mol Ag(s)

3. Convert 1.147 g of copper metal to number of moles:

  • n = mass in grams / atomic mass
  • n = 1.147 g / 63.546 g/mol = 0.01805 mol Cu (keep four signiticant digits such as the mass in grams is measured)

4. Use proportionality

1 mol Cu(s) / 2 mol Ag(s) = 1 mol Cu / 2 mol Ag = 0.01805 mol Cu / x

⇒ x = 0.01805 mol Cu × 2 mol Ag / 1 mol Cu = 0.03610 mol Ag.

5. Convert 0.03610 mol Ag to grams

  • mass in grams = number of moles × atomic mass of Ag
  • mass in grams = 0.03610 mols × 74.922 g/ mol = 2.705 g Ag (keep 4 significant digits)

Answer: 2.705 g Ag.

Question 3. A student reacts 1.147 g of copper with a solution containing excess silver nitrate and recovers 1.914 g of silver metal. Which reaction, a) or b), most likely occurred in the experiment? Explain your answer completely.

Answer: reaction b), most likely occurred in the experiment

Explanation:

  • As per equation a) (according to the answer for the question 1) by reacting 1.147 g of copper with excess of silver nitrate, the maximum recovery of silver metal, Ag (s) , is 1.3452 g

       So, it is not possible that a student may recover 1.914 g of silver metal from reaction a).

  • As per equation b) (according to the calculations performed to answer the question 2), 1.147 g of Cu(s) can yield (theoretical yield) 2.705 g (Ag (s)). So. it is most likely that reaction b) occurred in the experiment in which a student recovered 1.914 g of silver metal.

With this result, the percent yield is:

Percent yield = (actual yield / theoretical yield) × 100 = (1.914 g / 2.705 g) × 100 = 70.76%.

  • 70.76% yield is a likely result in an experiment, since the conditions almost never let to obtain the maximum (100%) recover.

Question 4. When copper metal and silver ion react, which reactant is oxidized and which is reduced?

Answer:  Copper is oxidized and silver is reduced

Explanation:

  • Oxidation is the loss of electrons yielding to an increase in the oxidation state (also called oxidation numbers).

  • Reduction is the gain of electrons yielding to a reduction in the oxidation state (oxidation number)

The net ionic equations show the changes in the oxidation numbers.

Take this one, Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) (reaction a)

Cu(s) has oxidation number 0, since it is alone, and the oxidation number of Cu²⁺(aq) is 2+, meaning that solid copper is the oxidized element.

2Ag⁺(aq) has oxidation number +1, and Ag(s), for the same reason above, has oxidation number 0; so the silver ion (2Ag⁺(aq)) was reduced.

Question 5. When copper metal and silver ion react, which is the more active metal?

Answer: copper metal is more active than silver ion.

Explanation:

The very fact that Cu(s) replaced Ag ion in the compound 2AgNO₃(aq) proves that copper metal is more active than silver ion: the more active elements are able to displace the less active elements in a compound.

When comparing metals, in a reduction-oxidation (redox) reaction , the reducing metal (the one that gets oxidized) is the most active metal.

Remember that the element that is oxidized, reduces the other element; so it is called the reducing agent. The reducing metal is the most active metal; in this case, this was the solid copper.

Normally, the metals are classified as per their activity in activity series. You can find the activity series in many tables. There, you will find that copper is more active than silver, confirming our conclusion.

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