The E values at 298 corresponding to the following two reduction electrode processes are i) Cu+/Cu = +0 52V ii) - Chemistry - Electrochemistry

Question

The E values at 298 corresponding to the following two reduction
electrode processes are i) Cu+/Cu = +0 52V ii) Cu2+/Cu = 0 16 V
Formulate the galvanic cell for their combination what will be the
cell potential? Calculate the del r G for the cell reaction
(E=96500 C/mol)

Vartika Jain · Accepted Answer

Dear Student,

Since anode electrode has less value of E°(reduction), so the galvanic cell obtained by the combination of anode and cathode will be :Cu+/Cu2+ ∥ Cu+/Cu(s)Since E°(reduction) value, +0
52>+0
16 V, so we haveCu+ + e- ↔ Cu      E°=+0
52 V     (cathode)Cu2+ + e- ↔ Cu+  E°= +0
16 V     (anode) Substracting the above two, 2Cu+ ↔ Cu + Cu2+     E°=0
52-0 16=0
36 VSince the reaction os a one electron change, n=1∆G°=-nFE°         =-1×96500×0
36         =-34540 J         =-34
74 kJ

The E values at 298 corresponding to the following two reduction electrode processes are i) Cu+/Cu = +0.52V ii) Cu2+/Cu = 0.16 V Formulate the galvanic cell for their combination. what will be the cell potential? Calculate the del r G for the cell reaction.(E=96500 C/mol)