Question

Помогите, пожалуйста подобрать коэффициенты ионно-электронным методом Cr2(SO4)3+Br2+NaOH=Na2CrO4+NaBr+Na2SO4+H2O SnCI2+K2Cr2O7+HCI=SnCI4+CrCI3+KCI+H2O CI2+HNO3=HCIO3+NO+H2O HCI+KMnO4=CI2+MnCI2+H2O+KCI PbO2+Mn(NO3)2+HNO3=Pb(NO3)2+HMnO4+H2O Pt+HNO3+HCI=PtCI4+NO+H2O K2Cr2O7+NaNO2+H2SO4=NaNO3+Cr2(SO4)3+K2SO4+H2O H2SO3+Br2+H2O=H2SO4+HBr Hg(NO3)2+H2O2+NaOH=Hg+O2+H2O+NaNO3

Answer 1

Cr2(SO4)3+Br2+12NaOH=2Na2CrO4+2NaBr+3Na2SO4+6H2O

Answer 2

Ion-electron method for balancing chemical equations

The ion-electron method, also known as the half-reaction method, is a technique used to balance chemical equations in redox reactions. It involves breaking down the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction. Each half-reaction is balanced separately, and then combined to give the balanced overall equation.

Let's go through the given chemical equations one by one and balance them using the ion-electron method.

1. Cr2(SO4)3 + Br2 + NaOH = Na2CrO4 + NaBr + Na2SO4 + H2O

To balance this equation, we need to identify the oxidation and reduction half-reactions.

Oxidation half-reaction: Cr2(SO4)3 → Na2CrO4 Reduction half-reaction: Br2 + 2e- → 2Br-

Balancing the oxidation half-reaction: Cr2(SO4)3 → 2CrO4^2- + 3SO4^2-

Balancing the reduction half-reaction: Br2 + 2e- → 2Br-

Now, we need to balance the number of electrons transferred in both half-reactions. In this case, 6 electrons are transferred in the oxidation half-reaction and 2 electrons in the reduction half-reaction.

To balance the electrons, we multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2:

2Cr2(SO4)3 → 4CrO4^2- + 6SO4^2- 3Br2 + 6e- → 6Br-

Now, we can combine the two half-reactions:

2Cr2(SO4)3 + 3Br2 + 6NaOH → 4Na2CrO4 + 6NaBr + 6Na2SO4 + 3H2O

2. SnCl2 + K2Cr2O7 + HCl = SnCl4 + CrCl3 + KCl + H2O

Oxidation half-reaction: SnCl2 → SnCl4 Reduction half-reaction: Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O

Balancing the oxidation half-reaction: SnCl2 → SnCl4

Balancing the reduction half-reaction: Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O

Balancing the electrons, we multiply the oxidation half-reaction by 6 and the reduction half-reaction by 2:

6SnCl2 → 6SnCl4 Cr2O7^2- + 14H+ + 12e- → 4Cr^3+ + 7H2O

Combining the two half-reactions:

6SnCl2 + Cr2O7^2- + 14H+ + 12HCl → 6SnCl4 + 4CrCl3 + KCl + 7H2O

3. Cl2 + HNO3 = HClO3 + NO + H2O + HCl

Oxidation half-reaction: Cl2 → HClO3 Reduction half-reaction: HNO3 → NO

Balancing the oxidation half-reaction: Cl2 + 2H2O → 2HClO3 + 2H+ + 2e-

Balancing the reduction half-reaction: HNO3 + 2H+ + 2e- → NO + 2H2O

Balancing the electrons, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1:

2Cl2 + 4H2O → 4HClO3 + 4H+ + 4e- 2HNO3 + 4H+ + 4e- → 2NO + 4H2O

Combining the two half-reactions:

2Cl2 + 2HNO3 + 4H+ → 4HClO3 + 2NO + 4H2O + 2HCl

4. KMnO4 + HCl = Cl2 + MnCl2 + H2O + KCl

Oxidation half-reaction: KMnO4 → MnCl2 Reduction half-reaction: HCl → Cl2

Balancing the oxidation half-reaction: 2KMnO4 + 16HCl → 2MnCl2 + 8H2O + 5Cl2 + 2KCl

Balancing the reduction half-reaction: 2HCl → Cl2 + 2H+ + 2e-

Combining the two half-reactions:

2KMnO4 + 18HCl → 2MnCl2 + 8H2O + 5Cl2 + 2KCl

5. PbO2 + Mn(NO3)2 + HNO3 = Pb(NO3)2 + HMnO4 + H2O

Oxidation half-reaction: PbO2 → Pb(NO3)2 Reduction half-reaction: Mn(NO3)2 → HMnO4

Balancing the oxidation half-reaction: PbO2 + 4HNO3 → Pb(NO3)2 + 2H2O + 4NO2

Balancing the reduction half-reaction: Mn(NO3)2 + 3H2O → HMnO4 + 2HNO3 + 2H+

Combining the two half-reactions:

PbO2 + Mn(NO3)2 + 4HNO3 + 3H2O → Pb(NO3)2 + HMnO4 + 2H2O + 4NO2 + 2HNO3 + 2H+

6. Pt + HNO3 + HCl = PtCl4 + NO + H2O

Oxidation half-reaction: Pt → PtCl4 Reduction half-reaction: HNO3 → NO

Balancing the oxidation half-reaction: Pt + 4Cl2 → PtCl4

Balancing the reduction half-reaction: HNO3 → NO

Combining the two half-reactions:

Pt + 4HNO3 + 4HCl → PtCl4 + NO + 3H2O

7. K2Cr2O7 + NaNO2 + H2SO4 = NaNO3 + Cr2(SO4)3 + K2SO4 + H2O

Oxidation half-reaction: K2Cr2O7 → Cr2(SO4)3 Reduction half-reaction: NaNO2 → NaNO3

Balancing the oxidation half-reaction: K2Cr2O7 + 7H2SO4 → Cr2(SO4)3 + K2SO4 + 7H2O + 3O2

Balancing the reduction half-reaction: 2NaNO2 + 2H2SO4 → 2NaNO3 + 2H2O + SO2

Combining the two half-reactions:

K2Cr2O7 + 7NaNO2 + 7H2SO4 → Cr2(SO4)3 + K2SO4 + 7H2O + 3O2 + 2NaNO3 + 2H2O + SO2

8. H2SO3 + Br2 + H2O = H2SO4 + HBr

Oxidation half-reaction: H2SO3 → H2SO4 Reduction half-reaction: Br2 → 2HBr

Balancing the oxidation half-reaction: H2SO3 + O2 → H2SO4

Balancing the reduction half-reaction: Br2 + 2H2O → 4HBr + O2

Combining the two half-reactions:

H2SO3 + Br2 + 2H2O → H2SO4 + 4HBr + O2

9. Hg(NO3)2 + H2O2 + NaOH = Hg + O2 + H2O + NaNO3

Oxidation half-reaction: Hg(NO3)2 → Hg Reduction half-reaction: H2O2 → O2

Balancing the oxidation half-reaction: Hg(NO3)2 + 2NaOH → Hg + 2NaNO3 + H2O

Balancing the reduction half-reaction: H2O2 → O2 + 2H+

Combining the two half-reactions:

Hg(NO3)2 + H2O2 + 2NaOH → Hg + O2 + H2O + 2NaNO3

Please note that the above-balanced equations are based on the ion-electron method.