in the reaction zn + h⁺ → zn²⁺ + h₂, zn is oxidized and h is reduced. how many electrons would be produced…

in the reaction zn + h⁺ → zn²⁺ + h₂, zn is oxidized and h is reduced. how many electrons would be produced and used in the balanced half - reactions? complete the balanced half - reactions: \n□h⁺+□e⁻→□h₂\n□zn→□zn²⁺+□e⁻

in the reaction zn + h⁺ → zn²⁺ + h₂, zn is oxidized and h is reduced. how many electrons would be produced and used in the balanced half - reactions? complete the balanced half - reactions: \n□h⁺+□e⁻→□h₂\n□zn→□zn²⁺+□e⁻

Answer

Explanation:

Step1: Analyze oxidation half - reaction

Zn loses electrons to form Zn²⁺. The oxidation state of Zn changes from 0 to +2. So, for 1 Zn atom, 2 electrons are produced. The oxidation half - reaction is $Zn\rightarrow Zn^{2 +}+2e^{-}$.

Step2: Analyze reduction half - reaction

H⁺ gains electrons to form H₂. The oxidation state of H changes from +1 to 0. For 1 H₂ molecule which has 2 H atoms, 2 H⁺ ions are needed and 2 electrons are used. The reduction half - reaction is $2H^{+}+2e^{-}\rightarrow H_{2}$.

Answer:

2 electrons are produced and used. $2H^{+}+2e^{-}\rightarrow H_{2}$ $Zn\rightarrow Zn^{2 +}+2e^{-}$