Calculate the expected voltages for the electrochemical cells made by connecting

Question

Calculate the expected voltages for the electrochemical cells made by connecting the copper(ll) half reaction to each of the others in the list above. Remember that one of the listed reductions must be reversed to become an oxidation for anything to happen and that when a half rxn is reversed, the sign of its potential is reversed. For the voltage to be a positive number (indicating a spontaneous reaction), Which half rxn will be the cathode in each pairing? an example L Ag^+ + e^- rightarrow Ag E degree = +.79v Cl_2 + 2e^- rightarrow 2 Cl^- E degree = + 1.36 v For a E_cell > 0, silver half rxn must be reversed to become an oxidation: Ag rightarrow ag^+ + e^- E degree = -.79 overall rxn is 2 Ag + Cl_2 rightarrow 2 Ag^+ + 2Cl^- E_cell = -.79 + 1.36 = +.57 v (Cl_2 electrode is cathode)

Explanation / Answer

the given data

Ag+   + e-   -------------> Ag E0 = +0.79 V

Cl2   + 2e-  ---------------> 2Cl-   E0 = + 1.36 V

for copper ion we have :-

Cu2+  + 2e-  -----------> Cu(s) E0 = +0.337 V

the two reactions that will happen will be :-

2Ag+   + 2 e-   -------------> 2Ag E0 = +0.79 V cathode

Cu(s)   -----------> Cu2+  + 2e-   E0 = - 0.337 V anode

E0cell = 0.79 V - 0.337 V = 0.453 V

the second reaction will be :-

Cl2   + 2e-  ---------------> 2Cl-   E0 = + 1.36 V Cathode

Cu(s)   -----------> Cu2+  + 2e-   E0 = - 0.337 V anode

E0cell = 1.36 V - 0.337 V = 1.023 V

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.