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What is the cell potential for the following electrochemical reaction at 25°C? Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)Given: E° (Zn²⁺/Zn) = -0.76 VE° (Cu²⁺/Cu) = +0.34 V

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To calculate the cell potential for the given electrochemical reaction, we need to find the difference in standard reduction potentials between the two half-reactions. The reaction is:Zn s  + Cu aq   Zn aq  + Cu s The half-reactions are:Zn s   Zn aq  + 2e  oxidation Cu aq  + 2e  Cu s   reduction Given standard reduction potentials are:E  Zn/Zn  = -0.76 VE  Cu/Cu  = +0.34 VSince Zn is being oxidized, we need to reverse the sign of its standard reduction potential to get the standard oxidation potential:E  Zn/Zn  = +0.76 VNow, we can calculate the cell potential  Ecell  using the formula:Ecell = Ecathode - EanodeIn this case, Cu/Cu is the cathode  reduction  and Zn/Zn is the anode  oxidation :Ecell =  +0.34 V  -  +0.76 V  = -0.42 VHowever, since the cell potential should be positive for a spontaneous reaction, we need to reverse the reaction:Zn aq  + Cu s   Zn s  + Cu aq Now the half-reactions are:Zn aq  + 2e  Zn s   reduction Cu s   Cu aq  + 2e  oxidation And the standard reduction potentials are:E  Zn/Zn  = -0.76 VE  Cu/Cu  = -0.34 VNow, we can calculate the cell potential  Ecell  using the formula:Ecell = Ecathode - EanodeIn this case, Zn/Zn is the cathode  reduction  and Cu/Cu is the anode  oxidation :Ecell =  -0.76 V  -  -0.34 V  = 0.42 VSo, the cell potential for the given electrochemical reaction at 25C is 0.42 V.

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