question 27 (1 point)
what is the change in oxidation number of an atom of chromium in the half - reaction (ce{cro_3^{2-}(aq) - cr_2o_3^{2-}(aq)})?
(\bigcirc) an decrease of 1
(\bigcirc) a increase of 1
(\bigcirc) an decrease of 2
(\bigcirc) a increase of 2
(\bigcirc) an decrease of 3

question 28 (1 point)
what is the change in oxidation number of an atom of sulfur in the half - reaction (ce{so_2(g) - so_4^{2-}(aq)})?
(\bigcirc) an decrease of 1
(\bigcirc) a increase of 1
(\bigcirc) an decrease of 2
(\bigcirc) a increase of 2
(\bigcirc) there is no change in oxidation number for sulfur

Question

question 27 (1 point)
what is the change in oxidation number of an atom of chromium in the half - reaction (ce{cro_3^{2-}(aq) -> cr_2o_3^{2-}(aq)})?
(\bigcirc) an decrease of 1
(\bigcirc) a increase of 1
(\bigcirc) an decrease of 2
(\bigcirc) a increase of 2
(\bigcirc) an decrease of 3

question 28 (1 point)
what is the change in oxidation number of an atom of sulfur in the half - reaction (ce{so_2(g) -> so_4^{2-}(aq)})?
(\bigcirc) an decrease of 1
(\bigcirc) a increase of 1
(\bigcirc) an decrease of 2
(\bigcirc) a increase of 2
(\bigcirc) there is no change in oxidation number for sulfur

Accepted Answer

### Question 27
# Explanation:
## Step1: Find oxidation number in $ \text{CrO}_5^{2-} $
Let oxidation number of Cr be $ x $. Oxygen (O) is -2 (except in peroxides, but here in $ \text{CrO}_5 $, there are peroxy bonds, but for simplicity, first assume O is -2). The charge of the ion is -2. So:
$ x + 5(-2) = -2 $
$ x - 10 = -2 $
$ x = 8 $? Wait, no, actually $ \text{CrO}_5 $ has 2 peroxy O (each -1) and 3 normal O (-2). Let's correct: Let number of peroxy O be 2 (so 2*(-1)) and 3*(-2). Then:
$ x + 2(-1) + 3(-2) = -2 $
$ x - 2 - 6 = -2 $
$ x - 8 = -2 $
$ x = 6 $.

## Step2: Find oxidation number in $ \text{Cr}_2\text{O}_7^{2-} $
Let oxidation number of Cr be $ y $. O is -2. Charge is -2.
$ 2y + 7(-2) = -2 $
$ 2y - 14 = -2 $
$ 2y = 12 $
$ y = 6 $? Wait, no, wait the half-reaction is $ \text{CrO}_5^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. Wait, maybe I made a mistake. Wait, actually, in $ \text{CrO}_5 $ (peroxochromate), the oxidation state of Cr is +6 (since 5 O: 4 as -1 (peroxy) and 1 as -2? Wait, no, formula: $ \text{CrO}_5 $ has structure with 2 peroxy groups (4 O at -1) and 1 O at -2. So:
$ x + 4(-1) + 1(-2) = -2 $ (since ion is $ \text{CrO}_5^{2-} $)
$ x - 4 - 2 = -2 $
$ x - 6 = -2 $
$ x = 4 $? Wait, I'm confused. Let's use another approach. Wait, the half-reaction: $ \text{CrO}_5^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. Let's balance Cr first. Multiply $ \text{CrO}_5^{2-} $ by 2: $ 2\text{CrO}_5^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. Now count O: left has 10 O, right has 7 O. So we need to add 3 O (or $ \text{H}_2\text{O} $, but maybe acidic/basic, but for oxidation number, let's find Cr in each.

Wait, maybe the correct way: In $ \text{CrO}_5^{2-} $, Cr is +6? No, wait, let's check the oxidation state change. Wait, the answer options are decrease/increase of 1,2,3. Let's re-express:

Wait, maybe the half-reaction is $ \text{CrO}_4^{2-} \to \text{Cr}_2\text{O}_7^{2-} $? No, the question is $ \text{CrO}_5^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. Wait, maybe a typo, but assuming the standard problem: Wait, no, let's do it properly.

Wait, in $ \text{CrO}_5^{2-} $: Let’s let Cr be $ x $, O: in peroxides, O is -1, and normal O is -2. $ \text{CrO}_5 $ has 4 O as -1 (two peroxy groups) and 1 O as -2. So:

$ x + 4(-1) + 1(-2) = -2 $
$ x - 4 - 2 = -2 $
$ x = 4 $.

In $ \text{Cr}_2\text{O}_7^{2-} $: Cr is $ y $, O is -2.

$ 2y + 7(-2) = -2 $
$ 2y -14 = -2 $
$ 2y = 12 $
$ y = 6 $. Wait, that can't be. Wait, no, the half-reaction is $ \text{CrO}_5^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. Let's balance Cr: 2 $ \text{CrO}_5^{2-} $ gives 2 Cr, so $ \text{Cr}_2\text{O}_7^{2-} $ has 2 Cr. Now O: left has 10 O, right has 7 O. So we need to add 3 O? No, maybe in acidic medium, but for oxidation number, let's see the change. Wait, maybe I messed up the oxidation state in $ \text{CrO}_5 $. Actually, $ \text{CrO}_5 $ (peroxochromic acid anhydride) has Cr in +6, because 5 O: 4 as -1 (peroxy) and 1 as -2: $ 6 + 4(-1) + 1(-2) = 0 $ (for neutral $ \text{CrO}_5 $), but as ion $ \text{CrO}_5^{2-} $, it's $ x + 4(-1) + 1(-2) = -2 $ → $ x -6 = -2 $ → $ x=4 $. Wait, this is confusing. Maybe the intended reaction is $ \text{CrO}_4^{2-} \to \text{Cr}_2\text{O}_7^{2-} $, but no. Wait, the options include "an increase of 1", etc. Wait, maybe the correct approach:

Wait, let's assume that in $ \text{CrO}_5^{2-} $, Cr is +6, and in $ \text{Cr}_2\text{O}_7^{2-} $, Cr is +6? No, that can't be. Wait, maybe the half-reaction is $ \text{CrO}_2^{2-} \to \text{Cr}_2\text{O}_7^{2-} $? No, the question says $ \text{CrO}_5^{2-} $.

Wait, maybe the answer is "an increase of 1"? No, let's check again. Wait, maybe I made a mistake. Let's look for similar problems. Actually, in $ \text{CrO}_5 $ (peroxochromate), the oxidation state of Cr is +6 (because 5 O: 4 at -1 (peroxy) and 1 at -2: 6 + 4*(-1) + 1*(-2) = 0 for neutral, but as ion $ \text{CrO}_5^{2-} $, it's 6 + 4*(-1) + 1*(-2) = 0, but ion charge is -2, so maybe my structure is wrong. Alternatively, maybe the problem has a typo, and it's $ \text{CrO}_4^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. In $ \text{CrO}_4^{2-} $, Cr is +6 (x + 4*(-2) = -2 → x=6). In $ \text{Cr}_2\text{O}_7^{2-} $, 2x + 7*(-2) = -2 → 2x=12 → x=6. No change. But that's not an option.

Wait, maybe the half-reaction is $ \text{CrO}_5^{2-} \to \text{Cr}^{3+} $? No, the question is $ \text{CrO}_5^{2-} \to \text{Cr}_2\text{O}_7^{2-} $. I think I'm overcomplicating. Let's check the options. The options are decrease of 1, increase of 1, decrease of 2, increase of 2, decrease of 3.

Wait, another approach: Let's calculate oxidation number of Cr in $ \text{CrO}_5^{2-} $:

Let Cr = x, O = -2 (ignoring peroxy for a moment, even though that's wrong, but maybe the problem assumes O is -2). Then:

x + 5*(-2) = -2 → x -10 = -2 → x=8. Impossible, since Cr max is +6. So O must be -1 (peroxy). Let's say 4 O are -1 (two peroxy groups) and 1 O is -2. Then:

x + 4*(-1) + 1*(-2) = -2 → x -4 -2 = -2 → x=4.

In $ \text{Cr}_2\text{O}_7^{2-} $, Cr is y: 2y + 7*(-2) = -2 → 2y=12 → y=6.

So change is 6 - 4 = +2. So increase of 2.

# Answer: a increase of 2

### Question 28
# Explanation:
## Step1: Oxidation number in $ \text{SO}_2 $
Let oxidation number of S be $ x $. O is -2.
$ x + 2(-2) = 0 $ (since $ \text{SO}_2 $ is neutral)
$ x - 4 = 0 $
$ x = 4 $.

## Step2: Oxidation number in $ \text{SO}_4^{2-} $
Let oxidation number of S be $ y $. O is -2. Charge is -2.
$ y + 4(-2) = -2 $
$ y - 8 = -2 $
$ y = 6 $.

## Step3: Calculate change
Change = 6 - 4 = +2. So increase of 2.

# Answer: a increase of 2

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QUESTION IMAGE

Question

Explanation:

Question 27

Step1: Find oxidation number in \( \text{CrO}_5^{2-} \)

Step2: Find oxidation number in \( \text{Cr}_2\text{O}_7^{2-} \)

Step1: Oxidation number in \( \text{SO}_2 \)

Step2: Oxidation number in \( \text{SO}_4^{2-} \)

Step3: Calculate change

Answer:

Question 28