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lutik1710 [3]
2 years ago
8

What is the smallest whole-number coefficient for h₂s when the equation h₂s mno₄⁻ → mn²⁺ so₄²⁻ is balanced in acidic solution?

Chemistry
1 answer:
Klio2033 [76]2 years ago
6 0

The lowest smallest whole-number coefficient for H₂S is 5. Option C

<h3>What is the smallest whole-number coefficient for H₂S?</h3>

Now we know that a redox reaction is one in which there is a loss and gain of electrons One specie is oxidized (looses electrons) while another specie is reduced (gains electrons). The electrons must be transferred leading to an increase in oxidation number of one specie and a decrease in the oxidation number of another.

Now the reduction half equation is;

MnO4^-(aq) + 8H^+(aq) + 5e^- ----> Mn^2+(aq) + 4H2O (l)

The oxidation  half equation is;

H2S(aq) ----> S^2-(aq) + 2H^+(aq) + 2e^-

Multiplying the reduction half equation by 2 and the  oxidation  half equation is 5

2MnO4^-(aq) + 16H^+(aq) + 10e^- ----> 2Mn^2+(aq) + 8H2O (l)

And;

5H2S(aq) ----> 5S^2-(aq) + 10H^+(aq) + 10e^-

The overall balanced reaction equation is;

2MnO4^-(aq) + 6H^+(aq) + 5H2S(aq) ----> 2Mn^2+(aq) + 8H2O (l) + 5S^2-(aq)

Thus the lowest smallest whole-number coefficient for H₂S is 5.

Learn more about redox reaction:brainly.com/question/13293425

#SPJ1

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State the oxidation number of S in <br><img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3
Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

  • The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
  • The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

<u><em>oxidation numbers of S= 4</em></u>

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

brainly.com/question/8990767

brainly.com/question/6498977

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