Question

question 1
in the reaction:
ca₍ₛ₎ + h₂o₍ₗ₎ ==> ca(oh)₂₍ₛ₎ + h₂₍₉₎
what is reduced and what is oxidized?
□ ca is reduced
□ hydrogen is reduced
□ ca is oxidized
□ oxygen is reduced
□ hydrogen is oxidized
□ oxygen is oxidized

Explanation:

Step1: Balance the reaction

First, balance the chemical equation. The unbalanced reaction is \( \text{Ca}_{(s)} + \text{H}_2\text{O}_{(l)}
ightarrow \text{Ca(OH)}_{2(s)} + \text{H}_2_{(g)} \). To balance it, we need 2 moles of \( \text{H}_2\text{O} \) to balance the oxygen and hydrogen: \( \text{Ca}_{(s)} + 2\text{H}_2\text{O}_{(l)}
ightarrow \text{Ca(OH)}_{2(s)} + \text{H}_2_{(g)} \).

Step2: Determine oxidation states

• For \( \text{Ca} \): In elemental form (\( \text{Ca}_{(s)} \)), oxidation state is 0. In \( \text{Ca(OH)}_2 \), calcium has an oxidation state of +2 (since \( \text{OH}^- \) is -1, and 2*(-1) = -2, so Ca is +2 to balance). So Ca goes from 0 to +2, which means it loses electrons (oxidation: loss of electrons, increase in oxidation state). Thus, Ca is oxidized.
• For H in \( \text{H}_2\text{O} \): In \( \text{H}_2\text{O} \), hydrogen has an oxidation state of +1. In \( \text{H}_2_{(g)} \), hydrogen is 0 (elemental form). So H goes from +1 to 0, which means it gains electrons (reduction: gain of electrons, decrease in oxidation state). Thus, hydrogen (in \( \text{H}_2\text{O} \)) is reduced.
• For O in \( \text{H}_2\text{O} \) and \( \text{Ca(OH)}_2 \): Oxidation state of O is -2 in both, so no change. So oxygen is neither oxidized nor reduced.

Step3: Evaluate the options

• "Ca is reduced": False, Ca is oxidized.
• "hydrogen is reduced": True, H in \( \text{H}_2\text{O} \) is reduced to \( \text{H}_2 \).
• "Ca is oxidized": True, Ca's oxidation state increases.
• "oxygen is reduced": False, O's oxidation state doesn't change.
• "hydrogen is oxidized": False, H's oxidation state decreases (reduction).
• "oxygen is oxidized": False, O's oxidation state doesn't change.

Answer:

The correct options are:

• hydrogen is reduced
• Ca is oxidized