Answer:
Explanation:
reduction potential of Ag+/Ag = 0.80 V
the oxidation potential of Br-/Br2 = -1.066 V
reduction potential of Na+/Na = 2.71 V
reduction potential of 2H+/H2 = 0.00 V
oxidation potential of 2F-/F2 = -2.87 V
oxidation potential of 2H2O↔ O2 + 4H+ + 4e- = -1.23 V
So, at cathode reduction occurs so, here Ag(s) will form because it has high reduction potential.
However, at anode oxidation occurs and , here the substance with high oxidation potential can oxidises easily so bromine will form.
At anode oxidation occurs and, here the substance with high oxidation potential can oxidize easily so bromine will form.
oxidation potential of 2H2O↔ O2 + 4H+ + 4e- = -1.23 V
When the reduction potential of Ag+/Ag = 0.80 V
Also the oxidation potential of Br-/Br2 is = -1.066 V
The reduction potential of Na+/Na is = 2.71 V
Then reduction potential of 2H+/H2 is = 0.00 V
After that oxidation potential of 2F-/F2 is = -2.87 V
Now, The oxidation potential of 2H2O↔ O2 + 4H+ + 4e- is = -1.23 V
So, When at cathode reduction occurs so, here Ag(s) will form because it has high reduction potential.
Then, at anode oxidation occurs, and also, when the substance with high oxidation potential can oxidize easily so bromine will form.
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