How to Calculate Eº for 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)?

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SUMMARY

The standard electrode potential (Eº) for the reaction 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s) can be calculated using the provided values. Given Eº for Cu⁺(aq) + e⁻ → Cu(s) is 0.52V and Eº for Cu⁺(aq) → Cu²⁺(aq) + e⁻ is -0.15V, the overall Eº for the reaction is derived from the equation: Eº = Eº(cathode) - Eº(anode). Thus, Eº for the reaction is 0.52V - (-0.15V) = 0.67V.

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Given E[tex]^{0}[/tex]=0,52V for

Cu[tex]^{+}[/tex](aq) + e[tex]^{-}[/tex] --> Cu(s)

Calculate E[tex]^{0}[/tex] for the following reaction at 25ºC, 1atm.

2Cu[tex]^{+}[/tex](aq) --> Cu[tex]^{2+}[/tex](aq) + Cu(s)

Any ideas?
 
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Is that all the data you're given? Don't you need the E[tex]^{0}[/tex] for Cu+ => Cu2+?
 


While doing these exercises we can check the standart potential reduction table.

E0 for Cu+ => Cu2+ + 1e- is -0,15V

What would you do then?
 

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