What is produced at each electrode in the electrolysis of an aqueous solution of both nabr and agf?

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.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

the answer is 12 as i may remember

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