Question

Расставьте коэффициенты методом электронного баланса. Укажите окислитель и восстановитель, процессы окисления и восстановления. S + H2SO4 ® SO2 + H2O

Answer 1

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Answer 2

Introduction

To balance the given chemical equation using the method of electronic balancing, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, we have the equation:

S + H2SO4 → SO2 + H2O

Balancing the Equation

Let's start by balancing the sulfur (S) atoms. We have one sulfur atom on the left side and one on the right side, so the sulfur atoms are already balanced.

Next, let's balance the hydrogen (H) atoms. We have two hydrogen atoms in the H2SO4 molecule on the left side, and two hydrogen atoms in the H2O molecule on the right side. Therefore, the hydrogen atoms are already balanced.

Now, let's balance the oxygen (O) atoms. We have four oxygen atoms in the H2SO4 molecule on the left side, and two oxygen atoms in the SO2 and H2O molecules on the right side. To balance the oxygen atoms, we can add another H2O molecule to the right side of the equation:

S + H2SO4 → SO2 + 2H2O

Now, the equation is balanced in terms of the number of atoms of each element.

Oxidation and Reduction

In the given equation, S is being oxidized from an oxidation state of 0 to +4 in SO2. Therefore, S is the reducing agent (or the substance that is being oxidized).

On the other hand, H2SO4 is being reduced from an oxidation state of +6 to +4 in SO2. Therefore, H2SO4 is the oxidizing agent (or the substance that is being reduced).

Summary

To summarize, the balanced equation is:

S + H2SO4 → SO2 + 2H2O

In this equation, S is the reducing agent, H2SO4 is the oxidizing agent, and the processes of oxidation and reduction occur.