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Fudgin [204]
3 years ago
13

For a particular redox reaction ClO2– is oxidized to ClO4– and Fe3 is reduced to Fe2 . Complete and balance the equation for thi

s reaction in basic solution. Phases are optional.
Please HELP, I don't know how to balance redox reactions in basic solutions...!
Chemistry
1 answer:
creativ13 [48]3 years ago
8 0
We have to balance the equation in basic medium:
ClO₂⁻ → ClO₄⁻
Chlorine atoms are balanced 
we will balance oxygen atoms by adding water to the side with less oxygen
2 H₂O + ClO₂⁻ → ClO₄⁻
Now balance hydrogens by adding H⁺ first
2 H₂O + ClO₂⁻ → 4 H⁺ + ClO₄⁻
The charge will be balanced by adding electrons to side where positive charge is more
2 H₂O + ClO₂⁻ →  4 e + 4 H⁺ + ClO₄⁻
Now balance H⁺ by adding same number of OH⁻ in both sides
2 H₂O + ClO₂⁻ + 4OH⁻ →  4 e + 4 H⁺ + 4OH⁻ + ClO₄⁻
H⁺ will neutralize OH⁻ to give water
2 H₂O + ClO₂⁻ + 4 OH⁻ → 4 e + 4 H₂O + ClO₄⁻
Balanced half reaction will be:
ClO₂⁻ + 4 OH⁻ → 4 e + 2 H₂O + ClO₄⁻ → (1)
Now balance the second half (reduction):
Fe⁺³ → Fe⁺²
Balance charge by adding electrons:
e + Fe⁺³ → Fe⁺² → (2) Multiply by 4 and add the two equations to get:
ClO₂⁻ + 4 OH⁻ + 4 Fe⁺³  → 4 Fe⁺² + 2 H₂O + ClO₄⁻
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Molybdenum can form a wide series of halide compounds, including four different fluoride compounds. The percent by mass of molyb
azamat

The formula and names of the compounds are:

1. Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride  

4. Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride  

1. Determination of the name and formula of the molybdenum fluoride having 63.0% of molybdenum.

Molybdenum (Mo) = 63.0%

Fluorine (F) = 100 – 63 = 37%

<h3>Formula =? </h3>

Mo = 63.0%

F = 37%

Divide by their molar mass

Mo = 63.0 / 96 = 0.656

F = 37 / 19 = 1.947

Divide by the smallest

Mo = 0.656 / 0.656 = 1

F = 1.947 / 0.656 = 3

Therefore,

Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Determination of the name and formula of the molybdenum fluoride having 56.0% of molybdenum.

Molybdenum (Mo) = 56.0%,

Fluorine (F) = 100 – 56 = 44%

<h3>Formula =? </h3>

Mo = 56%

F = 44%

Divide by their molar mass

Mo = 56 / 96 = 0.583

F = 44 / 19 = 2.316

Divide by the smallest

Mo = 0.583 / 0.583 = 1

F = 2.316 / 0.583 = 4

Therefore,

Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Determination of the name and formula of the molybdenum fluoride having 50.0% of molybdenum.

Molybdenum (Mo) = 50.0%,

Fluorine (F) = 100 – 50 = 50%

<h3>Formula =? </h3>

Mo = 50%

F = 50%

Divide by their molar mass

Mo = 50 / 96 = 0.520

F = 50 / 19 = 2.632

Divide by the smallest

Mo = 0.520 / 0.520 = 1

F = 2.632 / 0.520 = 5

Therefore,

Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride  

4. Determination of the name and formula of the molybdenum fluoride having 46.0% of molybdenum.

Molybdenum (Mo) = 46.0%,

Fluorine (F) = 100 – 46 = 54%

<h3>Formula =? </h3>

Mo = 46%

F = 54%

Divide by their molar mass

Mo = 46 / 96 = 0.479

F = 54 / 19 = 2.842

Divide by the smallest

Mo = 0.479 / 0.479 = 1

F = 2.842 / 0.479 = 6

Therefore,

Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride  

Learn more: brainly.com/question/11185156

7 0
3 years ago
What characteristics determine how easily two substances change temperature? Check all that apply.
Vinil7 [7]

amount of time the two substances are in contact. area in contact between the two substances. specific heat of the material that makes up the substances. the density of the two substances in contact.

4 0
3 years ago
You mix a blue and a clear liquid together and they turn bright green. Has a chemical reaction occurred?
Margarita [4]
Yes because color change is a sign of a chemical reaction.
7 0
3 years ago
Read 2 more answers
A sample of a compound contains 41.33 g of carbon and 8.67 g of hydrogen. The molar mass of the
mihalych1998 [28]

Answer:

C6H15

Explanation:

First, we calculate the empirical formula as follows:

C = 41.33 g

H = 8.67 g

We convert each mass value to mole by dividing each element by its molar mass (C = 12g/mol, H = 1g/mol)

C = 41.33g ÷ 12g/mol = 3.44mol

H = 8.67g ÷ 1g/mol = 8.67mol

Next, we divide each mole value by the smallest (3.44mol)

C = 3.44mol ÷ 3.44mol = 1

H = 8.67mol ÷ 3.44mol = 2.52

We multiply this ratio by 2 to get a simple whole number ratio

C = 1 × 2 = 2

H = 2.52 × 2 = 5.04

Based on this, the whole number ratio of C and H is 2:5, hence, the empirical formula is C2H5.

The molecular mass of the compound is given as 87.18 g/mol, hence, the molecular formula is calculated as follows:

(C2H5)n = 87.18

[12(2) + 1(5)]n = 87.18

[24 + 5]n = 87.18

(29)n = 87.18

n = 87.18 ÷ 29

n = 3.006

Approximately to whole number, n = 3

Hence, the molecular formula of the compound is [C2H5]3

= C6H15

4 0
3 years ago
Determine the possible traits of the calves if:
White raven [17]

Answer:

1. All red calves i.e. RR

2. All roan calves i.e RW

3. 2 red calves (RR) and two roan calves (RW)

Explanation:

According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):

1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR

2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW

3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).

7 0
3 years ago
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