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Calculate the standard enthalpy change of reduction for the following reaction: Cu2+(aq) + 2e- -> Cu(s) Given that the standard electrode potential of Cu2+(aq)/Cu(s) is +0.34 V and the standard electrode potential of Ag+(aq)/Ag(s) is +0.80 V.

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To calculate the standard enthalpy change of reduction for the given reaction, we can use the relationship between the standard electrode potential  E  and the standard Gibbs free energy change  G  as follows:G = -nFEwhere n is the number of moles of electrons transferred in the reaction  in this case, 2 , F is the Faraday constant  approximately 96,485 C/mol , and E is the standard electrode potential.First, we need to find the G for the reduction of Cu2+ aq  to Cu s  using the given standard electrode potential:G = -nFEG = - 2 mol  96,485 C/mol  +0.34 V G = -65,609.8 J/molNow, we can use the relationship between G, H  standard enthalpy change , and S  standard entropy change  to find H:G = H - TSAt standard conditions, the temperature  T  is 298.15 K. However, we don't have the value for S. In many cases, the entropy change is relatively small compared to the enthalpy change, so we can make an approximation by assuming that G  H. This is not always accurate, but it can give us a rough estimate of the standard enthalpy change of reduction:H  GH  -65,609.8 J/molTherefore, the approximate standard enthalpy change of reduction for the reaction Cu2+ aq  + 2e- -> Cu s  is -65,609.8 J/mol.

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