io₃⁻(aq) + hso₃⁻(aq) → i⁻(aq) + so₄²⁻(aq)
which of the following properly identifies the number of electrons transferred per entity?
select one:
6e⁻/s
4e⁻/i
2e⁻/s
3e⁻/i
clear my choice
question 27
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when the skeletal equation above is balanced using lowest whole numbers, the coefficient for each species is as follows:
io₃⁻(aq)
hso₃⁻(aq)
i⁻(aq)
so₄²⁻(aq)

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Accepted Answer

# Explanation:
## Step1: Determine oxidation states
- For $ IO_3^- $: Let oxidation state of $ I $ be $ x $. $ x + 3(-2)= -1 \Rightarrow x = +5 $. In $ I^- $, oxidation state of $ I $ is $ -1 $. So $ I $ is reduced (gains electrons). The change in oxidation state per $ I $ atom: $ +5 \to -1 $, change of $ 6 $ electrons (gain of 6 $ e^- $).
- For $ HSO_3^- $: Let oxidation state of $ S $ be $ y $. $ +1 + y + 3(-2)= -1 \Rightarrow y = +4 $. In $ SO_4^{2-} $, oxidation state of $ S $ is $ +6 $. So $ S $ is oxidized (loses electrons). The change in oxidation state per $ S $ atom: $ +4 \to +6 $, change of $ 2 $ electrons (loss of 2 $ e^- $).

## Step2: Balance electrons transferred
- To balance electrons, find the least common multiple of 6 (for $ I $) and 2 (for $ S $). LCM of 6 and 2 is 6.
- For reduction ( $ I $ ): Number of $ IO_3^- $ needed to gain 6 electrons: $ \frac{6}{6} = 1 $ (since each $ IO_3^- $ gains 6 $ e^- $). Wait, no: each $ I $ in $ IO_3^- $ gains 6 $ e^- $, so 1 $ IO_3^- $ gains 6 $ e^- $.
- For oxidation ( $ S $ ): Each $ HSO_3^- $ loses 2 $ e^- $, so to lose 6 $ e^- $, we need $ \frac{6}{2} = 3 $ $ HSO_3^- $ molecules.

## Step3: Balance the atoms and charges
- Reduction half - reaction: $ IO_3^- \to I^- $
  - Balance $ O $ by adding $ H_2O $: $ IO_3^- \to I^- + 3H_2O $
  - Balance $ H $ by adding $ H^+ $: $ IO_3^- + 6H^+ \to I^- + 3H_2O $
  - Balance charge by adding electrons: $ IO_3^- + 6H^+ + 6e^- \to I^- + 3H_2O $
- Oxidation half - reaction: $ HSO_3^- \to SO_4^{2-} $
  - Balance $ O $ by adding $ H_2O $: $ HSO_3^- + H_2O \to SO_4^{2-} $
  - Balance $ H $ by adding $ H^+ $: $ HSO_3^- + H_2O \to SO_4^{2-} + 3H^+ $
  - Balance charge by removing electrons: $ HSO_3^- + H_2O \to SO_4^{2-} + 3H^+ + 2e^- $
- Multiply oxidation half - reaction by 3 to balance electrons (since reduction gains 6 $ e^- $ and oxidation loses 2 $ e^- $ per half - reaction):
  - Oxidation (multiplied by 3): $ 3HSO_3^- + 3H_2O \to 3SO_4^{2-} + 9H^+ + 6e^- $
  - Reduction: $ IO_3^- + 6H^+ + 6e^- \to I^- + 3H_2O $
- Add the two half - reactions together:
  - $ IO_3^- + 6H^+ + 6e^-+3HSO_3^- + 3H_2O\to I^- + 3H_2O+3SO_4^{2-} + 9H^+ + 6e^- $
  - Cancel out common terms ( $ 6e^- $, $ 3H_2O $, and simplify $ H^+ $ ( $ 9H^+ - 6H^+ = 3H^+ $): $ IO_3^-+3HSO_3^-\to I^- + 3SO_4^{2-}+3H^+ $
  - If we consider basic medium (since the original reaction has $ HSO_3^- $ which can be in basic or acidic, but let's check the charges. Wait, maybe we made a mistake in medium. Let's try basic medium.
  - In basic medium, add $ OH^- $ to neutralize $ H^+ $. For the above equation, add 3 $ OH^- $ to both sides:
    - $ IO_3^-+3HSO_3^- + 3OH^-\to I^- + 3SO_4^{2-}+3H^+ + 3OH^- $
    - $ 3H^+ + 3OH^- = 3H_2O $, so $ IO_3^-+3HSO_3^- + 3OH^-\to I^- + 3SO_4^{2-}+3H_2O $
    - But let's go back to the electron transfer ratio. We know that the electron gain for $ I $ is 6 per $ IO_3^- $ and electron loss for $ S $ is 2 per $ HSO_3^- $. The ratio of $ IO_3^- $ to $ HSO_3^- $ from electron balance is 1:3.
    - Now, let's balance the $ I $ and $ S $ atoms first. From the electron balance, we have 1 $ IO_3^- $, 3 $ HSO_3^- $. Now, let's balance the $ I^- $ and $ SO_4^{2-} $. Since 1 $ IO_3^- $ produces 1 $ I^- $ and 3 $ HSO_3^- $ produces 3 $ SO_4^{2-} $. Now, let's check the charge balance.
    - Left - hand side: Charge of $ IO_3^- $ is - 1, charge of 3 $ HSO_3^- $ is $ 3\times(-1)= - 3 $, total charge=-4.
    - Right - hand side: Charge of $ I^- $ is - 1, charge of 3 $ SO_4^{2-} $ is $ 3\times(-2)= - 6 $, total charge=-7. Wait, this is a problem. We must have made a mistake in the half - reaction.
    - Let's re - do the oxidation half - reaction in basic medium.
    - Oxidation half - reaction: $ HSO_3^-\to SO_4^{2-} $
      - Balance $ O $: $ HSO_3^-+H_2O\to SO_4^{2-} $
      - Balance $ H $: $ HSO_3^-+H_2O\to SO_4^{2-}+3H^+ $ (in acidic), but in basic, add $ 3OH^- $ to both sides: $ HSO_3^-+H_2O + 3OH^-\to SO_4^{2-}+3H^+ + 3OH^- $
      - $ 3H^+ + 3OH^- = 3H_2O $, so $ HSO_3^-+4OH^-\to SO_4^{2-}+2H_2O $
      - Balance charge: $ HSO_3^-+4OH^-\to SO_4^{2-}+2H_2O + 2e^- $ (charge on left: - 1-4=-5; charge on right: - 2 + 0=-2; so we need to lose 3 electrons? Wait, no, earlier oxidation state calculation: $ S $ in $ HSO_3^- $ is +4, in $ SO_4^{2-} $ is +6. So the change is $ +4\to+6 $, so loss of 2 electrons. So the correct oxidation half - reaction in basic medium:
        - $ HSO_3^- \to SO_4^{2-} $
        - Balance $ O $: $ HSO_3^-+H_2O\to SO_4^{2-} $
        - Balance $ H $: $ HSO_3^-+H_2O\to SO_4^{2-}+3H^+ $
        - Add $ 3OH^- $ to both sides: $ HSO_3^-+H_2O + 3OH^-\to SO_4^{2-}+3H^+ + 3OH^- $
        - $ 3H^+ + 3OH^- = 3H_2O $, so $ HSO_3^-+4OH^-\to SO_4^{2-}+2H_2O $
        - Balance charge: Left side charge: $ - 1+(-4)=-5 $, right side charge: $ - 2+0=-2 $. So we need to lose 3 electrons? This means our earlier oxidation state calculation was wrong.
        - Wait, $ HSO_3^- $: $ H(+1), S, O(-2\times3 = - 6) $. So $ 1 + S-6=-1\Rightarrow S = +4 $. $ SO_4^{2-} $: $ S+(-8)=-2\Rightarrow S = +6 $. So the change is $ +4\to+6 $, loss of 2 electrons. So the charge on the left side of the oxidation half - reaction (before adding electrons) is $ - 1 $ (for $ HSO_3^- $) and on the right side is $ - 2 $ (for $ SO_4^{2-} $). So the difference in charge is $ (-2)-(-1)=-1 $. But we are losing electrons (oxidation), so the number of electrons lost is $ n $ such that $ -1 - n=-2\Rightarrow n = 1 $? No, this is confusing. Let's use the ion - electron method correctly.
        - Oxidation half - reaction ( $ S $ from +4 to +6, loss of 2 $ e^- $):
          - $ HSO_3^- \to SO_4^{2-} $
          - Add $ H_2O $ to balance $ O $: $ HSO_3^-+H_2O\to SO_4^{2-} $
          - Add $ H^+ $ to balance $ H $: $ HSO_3^-+H_2O\to SO_4^{2-}+3H^+ $
          - Add electrons to balance charge: $ HSO_3^-+H_2O\to SO_4^{2-}+3H^+ + 2e^- $ (Left charge: - 1, right charge: $ - 2+3 = +1 $. So to balance charge, we need to add 2 electrons to the right: $ HSO_3^-+H_2O\to SO_4^{2-}+3H^+ + 2e^- $ (charge on left: - 1, charge on right: $ - 2 + 3-2=-1 $)
        - Reduction half - reaction ( $ I $ from +5 to - 1, gain of 6 $ e^- $):
          - $ IO_3^- \to I^- $
          - Add $ H_2O $ to balance $ O $: $ IO_3^- \to I^-+3H_2O $
          - Add $ H^+ $ to balance $ H $: $ IO_3^-+6H^+\to I^-+3H_2O $
          - Add electrons to balance charge: $ IO_3^-+6H^++6e^-\to I^-+3H_2O $ (Left charge: $ - 1 + 6=+5 $, right charge: $ - 1+0=-1 $. So we add 6 electrons to the left: $ IO_3^-+6H^++6e^-\to I^-+3H_2O $ (charge on left: $ - 1 + 6-6=-1 $, charge on right: - 1)
        - Now, to balance electrons, multiply the oxidation half - reaction by 3 (to get 6 electrons lost) and the reduction half - reaction by 1 (to get 6 electrons gained):
          - Oxidation (×3): $ 3HSO_3^-+3H_2O\to 3SO_4^{2-}+9H^+ + 6e^- $
          - Reduction (×1): $ IO_3^-+6H^++6e^-\to I^-+3H_2O $
        - Add the two half - reactions:
          - $ 3HSO_3^-+3H_2O+IO_3^-+6H^++6e^-\to 3SO_4^{2-}+9H^+ + 6e^-+I^-+3H_2O $
          - Cancel out common terms: $ 3HSO_3^-+IO_3^-\to 3SO_4^{2-}+3H^++I^- $
        - Now, balance the charge. The left - hand side charge: $ 3\times(-1)+(-1)=-4 $. The right - hand side charge: $ 3\times(-2)+(-1)+3\times(+1)=-6 - 1 + 3=-4 $. So the charge is balanced. Now, let's check the $ H $ and $ O $ atoms.
          - Left - hand side $ H $: $ 3\times1 = 3 $ (from $ HSO_3^- $), right - hand side $ H $: 3 (from $ 3H^+ $). Left - hand side $ O $: $ 3\times3+3 = 12 $ (from $ 3HSO_3^- $ and $ IO_3^- $), right - hand side $ O $: $ 3\times4 = 12 $ (from $ 3SO_4^{2-} $). So the equation is balanced as $ IO_3^-+3HSO_3^-\to I^- + 3SO_4^{2-}+3H^+ $ (in acidic medium) or if we consider the original reaction has $ IO_3^- $, $ HSO_3^- $, $ I^- $, $ SO_4^{2-} $, the coefficients from the balanced equation (in acidic medium) are:
            - Coefficient of $ IO_3^- $: 1
            - Coefficient of $ HSO_3^- $: 3
            - Coefficient of $ I^- $: 1
            - Coefficient of $ SO_4^{2-} $: 3

# Answer:
- Coefficient of $ IO_3^-(aq) $: 1
- Coefficient of $ HSO_3^-(aq) $: 3
- Coefficient of $ I^-(aq) $: 1
- Coefficient of $ SO_4^{2-}(aq) $: 3

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QUESTION IMAGE

Question

Explanation:

Step1: Determine oxidation states

Step2: Balance electrons transferred

Step3: Balance the atoms and charges

Answer: