Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

question 33 (1 point) how many electrons are transferred in the followi…

Question

question 33 (1 point)
how many electrons are transferred in the following balanced redox reaction?
5no₂⁻(aq) + 6h⁺ + 2mno₄⁻(aq) → 5no₃⁻(aq) + 2mn²⁺(aq) + 3h₂o(el)
○2
○4
○6
○8
○10
question 34 (1 point)
how many hydroxide ions must be added to this reaction to balance in basic conditions? 5no₂⁻(aq) + 6h⁺ + 2mno₄⁻(aq) → 5no₃⁻(aq) + 2mn²⁺(aq) + 3h₂o(el)
○6 to the reactant side
○6 to the product side
○6 to both sides
○3 to the right side
○this reaction cannot be balanced in basic conditions

Explanation:

Response
Question 33

Step1: Determine oxidation states

For \( \text{NO}_2^- \) to \( \text{NO}_3^- \): N in \( \text{NO}_2^- \): let oxidation state be \( x \), \( x + 2(-2) = -1 \Rightarrow x = +3 \). N in \( \text{NO}_3^- \): \( x + 3(-2) = -1 \Rightarrow x = +5 \). Each N loses \( 5 - 3 = 2 \) electrons. There are 5 N atoms, so total loss: \( 5 \times 2 = 10 \) electrons.

For \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \): Mn in \( \text{MnO}_4^- \): \( x + 4(-2) = -1 \Rightarrow x = +7 \). Mn in \( \text{Mn}^{2+} \) is +2. Each Mn gains \( 7 - 2 = 5 \) electrons. There are 2 Mn atoms, so total gain: \( 2 \times 5 = 10 \) electrons.

Step2: Confirm electron transfer

Electrons lost = electrons gained = 10.

The reaction in acidic conditions has \( 6\text{H}^+ \) on the reactant side. To balance in basic conditions, add \( 6\text{OH}^- \) to both sides (to neutralize \( \text{H}^+ \) and form \( \text{H}_2\text{O} \)). \( 6\text{H}^+ + 6\text{OH}^-
ightarrow 6\text{H}_2\text{O} \). On the reactant side, we had \( 6\text{H}^+ \), so adding \( 6\text{OH}^- \) to reactant side (and product side, but the \( \text{H}^+ \) and \( \text{OH}^- \) form \( \text{H}_2\text{O} \) on reactant side, and adjust \( \text{H}_2\text{O} \) on product side). Wait, the original acidic reaction has \( 6\text{H}^+ \) (reactant). To go to basic, add \( 6\text{OH}^- \) to reactant side (to react with \( \text{H}^+ \) to make \( \text{H}_2\text{O} \)). So the number of \( \text{OH}^- \) added to reactant side is 6.

Answer:

10

Question 34