Electrolysis of AgF(aq) in acidic solution

In a voltaic cell,  the electrons emanate from the negative terminal(ANODE) to the external circuit. In this case the electrolytic cell is part of the external circuit connected to the voltage source. Thus, the electrode of the electrolytic cell that is connected to the negative terminal of the voltage source is the cathode of the cell; it receives electrons that are used to reduce a substance. The electrons that are removed during the oxidation process at the anode travel to the positive terminal of the voltage source, thus completing the circuit of the cell.

Electrolysis of AgF(aq) in acidic solution leads to the formation of silver metal and oxygen gas-

At Cathode, always reduction occurs. We observe that Ag+(aq) is being reduced to Ag(s), corresponding to the following half-reaction:


Notice that the standard reduction potential for the reduction of Ag+ is more positive than for the reduction of either H2O(l) to H2(g) ( = -0.83 V) or H+(aq) to H2(g)( = 0.0 V). The more positive the value of , the more favorable the reduction is. Thus, Ag+ is the most favorable species for reduction in the solution.

The possible anode half-reactions are the oxidation of F to F2 or the oxidation of H2O to O2. (Because the solution is acidic, the concentration of OH is expected to be small, so we won’t consider the oxidation of OH.) The problem states that O2(g) is produced, so the anode reaction is:


As noted above in our discussion of the electrolysis of NaF(aq), the reduction of H2O is more favorable than the reduction of F.

We can put the two electrode reactions together to produce the overall reaction:




The standard cell emf is


Because the cell emf is negative, an external emf of at least +0.43 V must be provided to force the electrolysis reaction to occur.



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