Respuesta :
Answer is: this is reduction.
Reduction half reaction: Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq).
Reduction is lowering of oxidation number because ions or element gain electrons and oxidation number is decreased (in this example from +4 to +2).
In oxido-reduction reaction, at least one element lose and one element gain electrons.
Oxidation reaction is increasing of oxidation number of atom, because element lost electrons in chemical reaction.
Reduction half reaction: Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq).
Reduction is lowering of oxidation number because ions or element gain electrons and oxidation number is decreased (in this example from +4 to +2).
In oxido-reduction reaction, at least one element lose and one element gain electrons.
Oxidation reaction is increasing of oxidation number of atom, because element lost electrons in chemical reaction.
Reduction term describes this reaction
Sn⁴⁺(aq) + 2e⁻ ⇒ Sn²⁺(aq)
Further explanation
Reduction oxidation reaction is a chemical reaction in which there are changes in oxidation numbers
The general formula for determining oxidation numbers:
- 1. Single element atomic oxidation = 0. For example Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0
Group IA (Li, Na, K, Rb, Cs, and Fr) +1
Group IIA (Be, Mg, Ca, Sr and Ba) +2
H atoms in compounds = +1, except metal hydride compounds (Hydrogens that bind to groups IA or IIA) oxidation numbers H = -1, for example LiH, MgH₂, etc.
- 2. O atoms in compounds = -2, except OF2 = + 2 and at peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.
- 3. The oxidation number in the uncharged compound = 0,
we complete the answer choices
Which term best describes this reaction?
(1) ionization (3) oxidation
(2) neutralization (4) reduction
On reaction
Sn⁴⁺(aq) + 2e⁻ ⇒ Sn²⁺(aq)
the Sn atom experiences an reduction reaction (decrease in oxidation number) from 4+ to 2+
Learn more
an oxidation-reduction reaction
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loses electrons in a chemical reaction
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Keywords: reduction, oxidation