In the reaction, lithium (Li) is oxidized to lithium hydroxide (LiOH), while sodium (Na) is reduced from sodium hydroxide (NaOH) to elemental sodium (Na). The oxidation state of oxygen (O) in both NaOH and LiOH remains -2 throughout the reaction. Therefore, the oxidation state of oxygen does not change.
Lithium atom become the cation Li+.
Sodium (Na) is a metal that tends to lose an electron to become a Na+ ion, resulting in oxidation. In this process, sodium goes from its neutral state to a more positively charged state, indicating oxidation.
This change is called oxidation (increase of ox. state) or reduction (decrease)
They involve either transfer of electrons, or a change in the oxidation state of some atoms involved.
In the reaction where Na is oxidized to Na+ in a chemical reaction, the oxidation state of Na changes from 0 to +1. This means that Na loses one electron and is oxidized.
The oxidation state of Na changes from 0 in Na to +1 in NaOH and then back to 0 in Na in the reaction. Na starts as a neutral metal atom, gains an electron in NaOH to have an oxidation state of +1, and then loses that electron to return to its neutral state in the final product.
Li(0) to Li(+1) when Na(+1) transforms to Na(0); 1 electron is transferred from Li(0) to Na(+1) in this redox reaction
The oxidation states for Na, Br, and O are +1, +5, and -2, respectively. In NaBrO3, there is 1 Na, 1 Br, and 3 O atoms, leading to a total charge of 0. Thus, the oxidation state of NaBrO3 is +5 to balance the charge.
Not sure what L(s) is supposed to be, but regardless, the oxidation state of Na changes from 1+ in NaOH to zero in Na(s).
The oxidation state of an element with oxidation state 0 cannot change, as it already has a balance of electrons.
+1 for each Na -1 for oxygen (as it is peroxide)
In Na2SO4, the oxidation state of sodium (Na) is +1, the oxidation state of sulfur (S) is +6, and the oxidation state of oxygen (O) is -2. To calculate the oxidation state of the whole compound, you can use the rule that the sum of the oxidation states in a neutral compound is zero, so in this case it would be +1*2 + (-2)*4 = 0.
The oxidation number of H in NaHSO4 is +1. In this compound, Na has an oxidation state of +1, S has an oxidation state of +6, and O has an oxidation state of -2. By adding up the oxidation states and solving for H, it is determined to be +1.
Sodium Oxide, or NaOH, has no oxidation state. It has a charge, which is zero. The elements that make up NaOH, however, do have oxidation states. The oxidation state of sodium (Na) is +1, and it will forever be +1 because it is impossible for it to be anything else, no matter what situation. The same applies for Hydrogen. Oxygen has an oxidation state of -2, and almost always will have an oxidation state of -2. There is one notable exception: H2O2. In this case, since the total charge of the compound is neutral and the oxidation state of Hydrogen must be +1 and, seeing as there are two hydrogens, bringing the overall charge up to 2, the oxidation state of oxygen must be -1. If it was -2, then the molecule would have an overall charge of -2.
The oxidation number of Na in NaNO3 is +1, since Na typically has a +1 oxidation state in compounds. The oxidation number of N in NO3 is +5, since oxygen is usually assigned a -2 oxidation state and there are three oxygen atoms bonded to nitrogen in NO3.
The oxidation state of Na in Na2O2 is +1, as each Na atom carries a +1 charge while the O atoms carry a -1 charge. In KO2, the oxidation state of K is +1, similar to Na2O2, as each K atom carries a +1 charge while the O atoms carry a -1/2 charge.