IO3{-} + SO32{-} = SO42{-} + I{-} balance this redox reaction in alkaline medium using half cell
Answers
Step 1. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.
MnO4- + SO32- → MnO2 + SO42-
Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.
a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).
Mn+7O-24- + S+4O-232- → Mn+4O-22 + S+6O-242-
b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.
O:S+4O-232- → S+6O-242- + 2e- (S)
R:Mn+7O-24- + 3e- → Mn+4O-22 (Mn)
c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).
O:S+4O-232- → S+6O-242- + 2e-
R:Mn+7O-24- + 3e- → Mn+4O-22
Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.
a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.
O:S+4O-232- → S+6O-242- + 2e-
R:Mn+7O-24- + 3e- → Mn+4O-22
b) Balance the charge. For reactions in a basic solution, balance the charge so that both sides have the same total charge by adding an OH- ion to the side deficient in negative charge.
O:S+4O-232- + 2OH- → S+6O-242- + 2e-
R:Mn+7O-24- + 3e- → Mn+4O-22 + 4OH-
c) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.
O:S+4O-232- + 2OH- → S+6O-242- + 2e- + H2O
R:Mn+7O-24- + 3e- + 2H2O → Mn+4O-22 + 4OH-
Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E1/2 value.
Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.
O:S+4O-232- + 2OH- → S+6O-242- + 2e- + H2O| *3
R:Mn+7O-24- + 3e- + 2H2O → Mn+4O-22 + 4OH-| *2
O: 3S+4O-232- + 6OH- → 3S+6O-242- + 6e- + 3H2O
R: 2Mn+7O-24- + 6e- + 4H2O → 2Mn+4O-22 + 8OH-
Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.
3S+4O-232- + 2Mn+7O-24- + 6OH- + 6e- + 4H2O → 3S+6O-242- + 2Mn+4O-22 + 6e- + 8OH- + 3H2O
Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.
3S+4O-232- + 2Mn+7O-24- + H2O → 3S+6O-242- + 2Mn+4O-22 + 2OH-
Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.
Answer:
Explanation:
IO₃⁻ + SO₃⁻² ⇄ S0₄⁻² + I⁻
Oxidation half reaction:
SO₃⁻²→SO₄⁻²
There is a loss of 2 electrons and this is why oxidation takes place.
Equalizing the number of Oxygen atoms on both sides:
SO₃⁻²+H₂O → SO₄⁻²
Now equalize the number of hydrogen atoms
SO₃⁻²+H₂O → SO₄⁻² +2H⁺
SO₃⁻²+H₂O+2е⁻ → SO₄⁻² +2H⁺ (Eq.1)
Reduction half:
IO₃⁻ → I⁻
IO₃⁻ → I⁻ + 3H₂O
IO₃⁻ +6H⁺ → I⁻ + 3H₂O
IO₃⁻ +6H⁺ → I⁻ + 3H₂O + 6е⁻ (Eq.2)
Multiplying eq.1 by 6 and eq.2 by 2.
(SO₃⁻²+H₂O+2е⁻ → SO₄⁻² +2H⁺ ) x6
6SO₃⁻²+6H₂O+12е⁻ → 6SO₄⁻² +12H⁺ (Eq.3)
( IO₃⁻ +6H⁺ → I⁻ + 3H₂O + 6е⁻ ) x2
2IO₃⁻ +12H⁺ → 2I⁻ + 6H₂O + 12е⁻ (Eq.4)
Adding both eq.3 and 4.
6SO₃⁻²+6H₂O+12е⁻ → 6SO₄⁻² +12H⁺
2IO₃⁻ +12H⁺ → 2I⁻ + 6H₂O + 12е⁻
6SO₃⁻²+2IO₃⁻ → 2I⁻ + 6SO₄⁻²
As both the sides have equal number of powers i.e 14 so the equation is balanced.