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galben [10]
2 years ago
9

Cu + O2 --Cu2O

Chemistry
1 answer:
aniked [119]2 years ago
7 0

Answer:

a. 4Cu + O₂ → 2Cu₂O

b. 266g Cu

c. 102g Cu₂O

Explanation:

Based on the balanced reaction:

a. 4Cu + O₂ → 2Cu₂O

<em>4 moles of Cu reacts per mole of oxygen to produce 2 moles of Cu₂O.</em>

<em />

b. To solve this question we need to find the moles of Cu₂O. With these moles and the balanced reaction we can find the moles of Cu and its mass:

<em>Moles Cu₂O -Molar mass: 143.09g/mol-</em>

300g Cu₂O * (1mol / 143.09g) = 2.097 moles Cu₂O

<em>Moles Cu:</em>

2.097 moles Cu₂O * (4mol Cu / 2mol Cu₂O) 4.193 moles Cu

<em>Mass Cu - Molar mass: 63.546g/mol-</em>

4.193 moles Cu * (63.546g / mol) = 266g Cu

c. In the same way, using PV = nRT we can find the moles of Oxygen. With the moles of oxygen we can find the moles of Cu₂O and its mass:

n = PV / RT

P = 1atm at STP, V = 8.0L, R = 0.082atmL/molK, T = 273.15K at STP:

n = 1atm*8.0L / 0.082atmL/molK*273.15K

n = 0.357 moles O₂

<em>Moles Cu₂O:</em>

0.357 moles O₂ * (2mol Cu₂O / 1mol O₂) = 0.714 moles Cu₂O

<em>Mass Cu₂O -Molar mass: 143.09g/mol-</em>

0.714 moles Cu₂O * (143.09g/mol) =

102g Cu₂O

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Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

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\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

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