If current is passed in an electrolytic cell containing sodium nitrate, NaNO3, dissolved in water, what are the products at the anode and the cathode?
If it's dissolved in water then it is:
NaNO3 -> Na+ + NO3- right?
So you have...
Anode: NaNO3 -> Na+ + e-
Cathode : NaNO3 + e- -> NO3-
Is this correct?
Am I correct?
Yes and no.
Yes, you have Na^+ and NO3^-.
No, the products are not Na and NO3^-
H2O is easier to electrolyze than Na and NO3^-. The NaNO3 serves as an electrolyte to conduct electrons from electrode to electrode but the NaNO3 stays intact as the ions.
Yes, you are correct about the dissociation of NaNO3 in water. When NaNO3 dissolves in water, it dissociates into Na+ ions and NO3- ions.
In an electrolytic cell, during the electrolysis process, the positive electrode is called the anode, and the negative electrode is called the cathode.
At the anode (positive electrode), oxidation takes place. In this case, since you have Na+ ions available, the most likely product would be the oxidation of sodium ions. The equation for the reaction at the anode would be:
Anode:
2Na+ (from NaNO3) → 2Na+ + 2e-
This reaction releases two electrons, therefore increasing the charge of the Na+ ion to Na2+.
At the cathode (negative electrode), reduction takes place. In this case, the most likely reduction would be the reduction of water molecules due to the presence of NO3- ions. The equation for the reaction at the cathode would be:
Cathode:
2H2O + 2e- → 2OH- + H2(g)
In this reaction, water molecules (H2O) are reduced by gaining two electrons to form hydroxide ions (OH-) and hydrogen gas (H2).
So, the products at the anode would be Na2+ ions, and the products at the cathode would be hydroxide ions (OH-) and hydrogen gas (H2).