QuestionJuly 11, 2025

Identify the element being reduced and the element being oxidized in the reaction below. In each case, enter the element symbol. If the reaction isn't actually a redox reaction, enter NOPE for each answer. H_(2)SO_(4)+2NaOHarrow Na_(2)SO_(4)+2H_(2)O element being oxidized: element being reduced:

Identify the element being reduced and the element being oxidized in the reaction below. In each case, enter the element symbol. If the reaction isn't actually a redox reaction, enter NOPE for each answer. H_(2)SO_(4)+2NaOHarrow Na_(2)SO_(4)+2H_(2)O element being oxidized: element being reduced:
Identify the element being reduced and the element being oxidized in the reaction below.
In each case, enter the element symbol. If the reaction isn't actually a redox reaction,
enter NOPE for each answer.
H_(2)SO_(4)+2NaOHarrow Na_(2)SO_(4)+2H_(2)O
element being oxidized:
element being reduced:

Solution
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Answer

NOPE ### NOPE Explanation 1. Determine Oxidation States Assign oxidation states to each element in the reactants and products. In H_2SO_4, H is +1, O is -2, and S is +6. In NaOH, Na is +1, O is -2, and H is +1. In Na_2SO_4, Na is +1, O is -2, and S is +6. In H_2O, H is +1 and O is -2. 2. Compare Oxidation States Compare the oxidation states of elements on both sides of the equation. No changes in oxidation states are observed for any element. 3. Identify Redox Reaction Since there are no changes in oxidation states, this reaction is not a redox reaction.

Explanation

1. Determine Oxidation States<br /> Assign oxidation states to each element in the reactants and products. In $H_2SO_4$, H is +1, O is -2, and S is +6. In $NaOH$, Na is +1, O is -2, and H is +1. In $Na_2SO_4$, Na is +1, O is -2, and S is +6. In $H_2O$, H is +1 and O is -2.<br /><br />2. Compare Oxidation States<br /> Compare the oxidation states of elements on both sides of the equation. No changes in oxidation states are observed for any element.<br /><br />3. Identify Redox Reaction<br /> Since there are no changes in oxidation states, this reaction is not a redox reaction.
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