dilute aqueous solution of sodium fluoride is electrolyesed the product at rhe anod and cathod are
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At cathode:
Hydrogen is less reactive than Potassium, it gains one electron from the cathode and is discharged as hydrogen gas. Potassium ions remain in the solution.
The equation for reaction at the cathode is:
H+ + e- → H2
At anode:
O2 + 4H+ + 4e- -----> 2H2O (+1.23 V)
F2 + 2e- -----> 2F- (+2.87 V)
Oxidation of water is easier than Fluoride and hence the products of electrolysis are oxygen and hydrogen. and this gives hydrogen at cathode and oxygen at anode
Hydrogen is less reactive than Potassium, it gains one electron from the cathode and is discharged as hydrogen gas. Potassium ions remain in the solution.
The equation for reaction at the cathode is:
H+ + e- → H2
At anode:
O2 + 4H+ + 4e- -----> 2H2O (+1.23 V)
F2 + 2e- -----> 2F- (+2.87 V)
Oxidation of water is easier than Fluoride and hence the products of electrolysis are oxygen and hydrogen. and this gives hydrogen at cathode and oxygen at anode
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