calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?
1 answer:
The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.
<u>Explanation:</u>
An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.
So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to
ions. This process of acceptance of electrons is termed as reduction.
You might be interested in
Answer:
The answer to your question is Molarity = 0.0708
Explanation:
Data
NaOH 0.05 M Volume 1 = 3.87 ml Volume 2 = 25.11 ml
HCl 15 ml
Process
1.- Find the volume used of NaOH
25.11 - 3.87 = 21.24 ml = 0.02124 l
2.- Write the balanced equation of the reaction
NaOH + HCl ⇒ NaCl + H₂O
3.- Calculate the moles of NaOH in the solution
Molarity =
moles = Molarity x volume
moles = 0.05 x 0.02124
moles = 0.001062
4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.
1 mol of NaOH ------------- 1 mol of HCl
0.001062 moles of NaOH ------------ x
x = (0.001062 x 1) / 1
x = 0.001062 moles of HCl
5.- Find the molarity of HCl
Molarity =
Molarity = 0.0708