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Balance the following equations by using ion – electron method in basic solutionrn1. MnO4- + S2O3-2 s4o6-2 + Mn+2
Accepted Answer
Sure, here is the balanced equation in basic solution using the ion electron method:
1.
Reactants:
MnO4- + S2O3-2
Products:
S4O6-2 + Mn+2
Steps:
1. Write the unbalanced equation in basic solution:
MnO4- + S2O3-2 + H2O → S4O6-2 + Mn+2 + OH-
2. Assign oxidation numbers to each atom:
Reactants:
Mn (+7) O (-2)4 S (+3)2 O (-2)3
Products:
S (+6)4 O (-2)6 Mn (+2) O (-2) H (+1)
3. Identify the species that is oxidized and the species that is reduced:
Oxidized:
S (+3)2 → S (+6)4
Reduced:
Mn (+7) → Mn (+2)
4. Write the half-reaction for the oxidation:
2S2O3-2 → S4O6-2 + 2e-
5. Write the half-reaction for the reduction:
MnO4- + 4H2O + 3e- → MnO2 + 8OH-
6. Balance the half-reactions:
Oxidation:
2S2O3-2 → S4O6-2 + 2e-
Reduction:
MnO4- + 4H2O + 3e- → MnO2 + 8OH-
7. Multiply the half-reactions so that the number of electrons lost is equal to the number of electrons gained:
Oxidation:
5 x (2S2O3-2 → S4O6-2 + 2e-)
Reduction:
3 x (MnO4- + 4H2O + 3e- → MnO2 + 8OH-)
8. Add the balanced half-reactions together and cancel out any species that appear on both sides of the equation:
10S2O3-2 + 3MnO4- + 12H2O → 5S4O6-2 + 3MnO2 + 24OH-
9. Check the equation to make sure that it is balanced:
Reactants:
10S (+3)2 O (-2)3 + 3Mn (+7) O (-2)4 + 12H (+1) O (-2)
Products:
5S (+6)4 O (-2)6 + 3Mn (+4) O (-2)2 + 24H (+1) O (-2)
The equation is balanced!
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