Solve "2Cu+(aq) → Cu + Cu2+(aq)" Equation ΔrH | Exam Prep Help

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SUMMARY

The discussion focuses on calculating the heat of reaction (ΔrH) for the equation 2Cu+(aq) → Cu + Cu2+(aq). The provided half-reactions are Cu(s) → Cu+(aq) + e (ΔH = 602 kJ mol-1) and Cu(s) → Cu2+(aq) + 2e (ΔH = 795 kJ mol-1). By subtracting these equations, the ΔrH is determined to be 193 kJ mol-1. Additionally, the discussion emphasizes the importance of connecting the half-reactions for Cu+ and Cu2+ to accurately balance the reaction and understand the stability of the ions.

PREREQUISITES
  • Understanding of thermodynamics, specifically enthalpy changes.
  • Familiarity with half-reaction methods in electrochemistry.
  • Knowledge of oxidation and reduction processes.
  • Basic proficiency in chemical equation balancing.
NEXT STEPS
  • Study the concept of enthalpy changes in chemical reactions.
  • Learn about half-reaction methods in electrochemistry.
  • Research the stability of transition metal ions, focusing on Cu+ and Cu2+.
  • Explore the principles of oxidation and reduction in redox reactions.
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Ryan thermo
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Hi, I have my exam in a couple of weeks. And I get the jist of it. However their is one question that I am completely stuck on. Any help would be appreciated.

Looking over a last years practise text for exam and I am having trouble with this.

Cu(s) → Cu+(aq) + e ΔH = 602 kJ mol-1
Cu(s) → Cu2+(aq) + 2e ΔH = 795 kJ mol-1
Work out the Heat of Reaction for the following equation and comment on
the expected stabilities of Cu+ and Cu2+ ions as predicted from your
spdf answers in (d).

2Cu+(aq) → Cu + Cu2+(aq) ΔrH = ?
 
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Try to break it into two equations. Cu+ to Cu2+ and Cu+ to Cu.

For the first reaction, you can find its ΔH by subtracting the equations given in data. ΔH would be 193 KJmol-1
 
Alternative approach, probably simpler:

The problem you are having is that you do not have a half-reaction connecting Cu(+) and Cu(2+)

So why not reduce both mole of Cu(+) to Cu, and couple it with oxidizing one mole of Cu all the way to Cu(2+)? That will balance the electrons, and produce the exact overall effect that you are wanting.
 

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