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10 days ago

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4) Balance the following redox reaction in an acidic solution. What are the coefficients in front of H⁺ and Fe3+ in the balanced

reaction? How many moles of electrons are transferred? Fe2+(aq) + MnO4⁻(aq) → Fe3+(aq) + Mn2+(aq)

1 answer:

eduard [944]10 days ago

3 0

Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5 respectively.

- A total of 5 moles of electrons are exchanged.

Explanation:

This reaction is represented as:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Analyzing the oxidation states:

Fe²⁺ transitions to Fe³⁺

This indicates an increase in oxidation state → OXIDATION

Meanwhile, Mn in MnO₄⁻ starts with +7 and transforms into Mn²⁺

This suggests a decrease in oxidation state → REDUCTION

Let's formulate the half reactions:

Fe²⁺ → Fe³⁺ + 1e⁻ (it loses 1 mole of electrons)

MnO₄⁻ + 5e⁻ → Mn²⁺ (it gains 5 moles of electrons)

Next, we will balance the oxygen atoms. In an acidic environment, water is added to balance the oxygens on the opposite side. Since there are 4 oxygens on the reactant side, we add 4 H₂O to the product side.

MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

Now, to balance the hydrogen atoms, we have 8 hydrogens in the products, necessitating the inclusion of 8H⁺ in the reactants, yielding the complete half-reaction:

8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

Notably, there's 1e⁻ in the oxidation and 5e⁻ in the reduction. To cancel electrons, we must multiply the oxidation half-reaction by 5.

(Fe²⁺ → Fe³⁺ + 1e⁻) x 5

5Fe²⁺ → 5Fe³⁺ + 5e⁻

8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

By adding both half reactions, we have:

5Fe²⁺ + 8H⁺ + MnO₄⁻ + 5e⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O

The electrons cancel out, resulting in the balanced equation:

5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O

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A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temp

lorasvet [946]

Response: The increase in temperature is $0.53^0C$

Reasoning:

The amount of thermal energy needed to elevate the temperature of a given substance by one degree Celsius is referred to as the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat gained by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g (1kg=1000g)

c = heat capacity of water = $4.18J/g^0C$

Initial temperature = $T_i$

Final temperature = $T_f$

Temperature change , $\Delta T=T_f-T_i=?$

Substituting the values, we obtain:

$5280J=2400g\times 4.18J/g^0C\times \Delta T$

$\Delta T=0.53^0C$

Therefore, the temperature increase is $0.53^0C$

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14 days ago

A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.

Alekssandra [962]

Answer:

The right answer is "1.0100".

Explanation:

Assuming the total volume of the mixture is 100 ml.

Thus,

The volume of DMSO will be 10 mL and the volume of water will be 90 mL.

For DMSO:

= $10\times 1.1004$

= $11.004 \ g$

The total mass of the mixture will be:

= $90+11.004$

= $101.004 \ g$

Calculating the density of the mixture:

= $\frac{Mass}{Volume}$

= $\frac{101.004}{100}$

= $1.01004 \ g/mL$

Thus,

The specific gravity of the mixture is:

= $\frac{Density \ of \ mixture}{Density \ of \ water}$

= $\frac{1.01004}{1}$

= $1.0100$

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6 days ago

A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes combustion, the e

castortr0y [918]

Response:

9.606 g

Clarification:

Step 1: Write the balanced combustion equation

C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the moles for 11.27 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

11.27 g × (1 mol/18.02 g) = 0.6254 mol

Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O

The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol

Step 4: Calculate the mass for 0.2085 moles of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

0.2085 mol × 46.07 g/mol = 9.606 g

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10 days ago

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

KiRa [965]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × $\frac{1mol}{159,609g}$ = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × $\frac{1mol}{18,02g}$ = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

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8 days ago

Acetone major species present when dissolved in water

Alekssandra [962]

The compound is acetone ( CH₃-CO-CH₃)

Explanation:

1) Acetone is represented as CH₃-CO-CH₃.

2) This is a molecule formed by covalent bonds.

3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.

In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.

This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).

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11 days ago