Use the ion-electron method to write and balance the net ionic equation for each of the following reactions.
MnO41- + Sn(OH)31- MnO2 + Sn(OH)62- in basic solution
MnO41- + (COOH)2(aq) Mn2+ + CO2(g) in acidic solution
I2(aq) + S2O32- I1- + S4O62-
Zn(s) + HgO(s) ZnO(s) + Hg(l) in basic solution
Fe(s) + O2(g) Fe(OH)2 in basic solution
Thank you!!
To write and balance net ionic equations using the ion-electron method, follow these steps:
Step 1: Write the balanced chemical equation.
Step 2: Identify the spectator ions in the equation.
Step 3: Write the ions that participate in the chemical reaction as half-reactions.
Step 4: Balance the atoms and charges in each half-reaction.
Step 5: Multiply each half-reaction by the necessary coefficients to balance the charges.
Step 6: Combine the half-reactions and cancel out the common species to obtain the net ionic equation.
Let's apply these steps to each of the given reactions:
1) MnO4^-1 + Sn(OH)3^-1 -> MnO2 + Sn(OH)6^-2 in basic solution
Step 1: The balanced chemical equation is:
2MnO4^-1 + 3Sn(OH)3^-1 -> 2MnO2 + 3Sn(OH)6^-2
Step 2: In basic solution, OH^- is a spectator ion.
Step 3: Split the reaction into two half-reactions:
Half-reaction 1: MnO4^-1 -> MnO2
Half-reaction 2: Sn(OH)3^-1 -> Sn(OH)6^-2
Step 4: Balance the atoms and charges in each half-reaction:
Half-reaction 1: 4H2O + MnO4^-1 -> MnO2 + 8OH^-
Half-reaction 2: 5OH^- + 3Sn(OH)3^-1 -> Sn(OH)6^-2
Step 5: Multiply each half-reaction by the necessary coefficients to balance charges:
Multiply the first half-reaction by 3 and the second half-reaction by 2:
12H2O + 3MnO4^-1 -> 3MnO2 + 24OH^-
10OH^- + 6Sn(OH)3^-1 -> 6Sn(OH)6^-2
Step 6: Combine the half-reactions and cancel out common species:
12H2O + 3MnO4^-1 + 10OH^- + 6Sn(OH)3^-1 -> 3MnO2 + 24OH^- + 6Sn(OH)6^-2
Simplify:
3MnO4^-1 + 6Sn(OH)3^-1 -> 3MnO2 + 6Sn(OH)6^-2
2) MnO4^-1 + (COOH)2(aq) -> Mn^2+ + CO2(g) in acidic solution
Follow the same steps as in the previous example to obtain the net ionic equation.
3) I2(aq) + S2O3^2- -> I^- + S4O6^2-
Follow the same steps as in the first example to obtain the net ionic equation.
4) Zn(s) + HgO(s) -> ZnO(s) + Hg(l) in basic solution
Follow the same steps as in the first example to obtain the net ionic equation.
5) Fe(s) + O2(g) -> Fe(OH)2 in basic solution
Follow the same steps as in the first example to obtain the net ionic equation.
Please note that the net ionic equations provided above were obtained following the ion-electron method.