Question

Уравняйте методом электронного балансаMnO2+NaBr+H2SO4=Br2+MnSO4+ Na2SO4+H2OЗнаю,что бром отдает 2e- , Mn их получает . Получается такое уравнение:2NaBr + 2H2SO4 + MnO2 = Br2 + Na2SO4 + MnSO4 + 2H2O Mn(4+)+2e- = Mn(2+)2Br(-) - 2e- = Br2 Как подобрать коэффициенты?Распишите, пожалуйста, сам процесс подбора коэффициентов :)

Answer 1

MnO2 + 2 NaBr + 2 H2SO4 = Br2 + MnSO4 + Na2SO4 + 2 H2O
Mn(+4) + 2e = Mn(+2)/1/окислитель, восстановился
2Br(-) - 2e = Br2(0)/1/восстановитель, окислился
Я бы уравнивала, исходя из того, что перед марганцем не должно быть коэффициента, но, с другой стороны, перед бромидом натрия сразу нужно тыкнуть 2...а там уже по кислороду смотреть.

Answer 2

Balancing the Chemical Equation using the Electron Balance Method

To balance the chemical equation MnO2 + NaBr + H2SO4 = Br2 + MnSO4 + Na2SO4 + H2O using the electron balance method, we need to ensure that the number of electrons gained is equal to the number of electrons lost.

Here's a step-by-step process to balance the equation:

Step 1: Identify the oxidation states of each element in the equation.

- Mn in MnO2 has an oxidation state of +4. - Na in NaBr has an oxidation state of +1. - Br in NaBr and Br2 has an oxidation state of -1. - H in H2SO4 has an oxidation state of +1. - S in H2SO4 and Na2SO4 has an oxidation state of +6. - O in MnO2 and H2SO4 has an oxidation state of -2. - O in H2O has an oxidation state of -2.

Step 2: Determine the half-reactions for the oxidation and reduction processes.

In this equation, Mn is reduced from +4 to +2, gaining 2 electrons, while Br is oxidized from -1 to 0, losing 1 electron. Therefore, we have the following half-reactions:

Reduction half-reaction: MnO2 + 4H+ + 2e- → Mn2+ + 2H2O Oxidation half-reaction: 2Br- → Br2 + 2e-

Step 3: Balance the atoms in each half-reaction.

In the reduction half-reaction, we have 1 Mn atom on the left side and 1 Mn atom on the right side. The oxygen atoms are already balanced, and the hydrogen atoms are balanced with 4 on both sides.

In the oxidation half-reaction, we have 2 Br atoms on the left side and 2 Br atoms on the right side.

Step 4: Balance the charges in each half-reaction.

In the reduction half-reaction, we have a charge of +4 on the left side and a charge of +2 on the right side. To balance the charges, we need to add 2 electrons to the right side.

In the oxidation half-reaction, we have a charge of -1 on the left side and a charge of 0 on the right side. The charges are already balanced.

Step 5: Multiply the half-reactions by appropriate coefficients to balance the number of electrons.

To balance the number of electrons, we need to multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1.

2(MnO2 + 4H+ + 2e-) → 2(Mn2+ + 2H2O) 2(2Br- → Br2 + 2e-)

Step 6: Combine the balanced half-reactions to form the balanced overall equation.

Adding the two balanced half-reactions together, we get:

2MnO2 + 8H+ + 4Br- → 2Mn2+ + 4H2O + Br2

Step 7: Balance the remaining atoms in the equation.

Now, let's balance the remaining atoms in the equation.

On the left side, we have 2 Mn atoms, 8 H atoms, and 4 Br atoms. On the right side, we have 2 Mn atoms, 8 H atoms, and 2 Br atoms. To balance the Br atoms, we need to multiply the Br2 molecule by 2.

2MnO2 + 8H+ + 4Br- → 2Mn2+ + 4H2O + 2Br2

Step 8: Balance the charges in the equation.

The charges are already balanced in the equation.

Step 9: Balance the number of atoms on both sides of the equation.

On the left side, we have 2 Mn atoms, 8 H atoms, and 4 Br atoms. On the right side, we have 2 Mn atoms, 8 H atoms, and 4 Br atoms. The equation is now balanced.

Balanced Equation:

The balanced equation for the reaction MnO2 + NaBr + H2SO4 = Br2 + MnSO4 + Na2SO4 + H2O is:

2MnO2 + 8H+ + 4Br- → 2Mn2+ + 4H2O + 2Br2

Please note that this is a balanced equation using the electron balance method.