Question

Balance the redox reaction occurring in acidic aqueous solution:

S2O3 (2-) (aq) + Cl2 (g) = SO4 (-2) (aq) + Cl (-) (aq)

Answer 1

To balance the redox equation in acidic aqueous solution, follow a step-by-step process including splitting the reaction into half-reactions, balancing atoms, adding H+ and H2O, balancing charges, multiplying by coefficients, and adding the balanced half-reactions together.

Step-by-step explanation:

In order to balance the redox equation in acidic aqueous solution, you need to follow a step-by-step process:

1. Split the reaction into two half-reactions: one for oxidation and one for reduction.
2. Balance the atoms in each half-reaction, excluding H and O.
3. Balance the oxygens by adding H2O to the side that needs more O atoms. Balance the hydrogens by adding H+ to the side that needs more H atoms.
4. Balance the charges by adding electrons (e-) to the side that has a higher positive charge.
5. Multiply the half-reactions by appropriate coefficients so that the electrons cancel out.
6. Add the balanced half-reactions together and simplify if necessary.

For the given redox reaction:

S2O3 (2-) (aq) + Cl2 (g) = SO4 (-2) (aq) + Cl (-) (aq)

The balanced equation in acidic aqueous solution is:

S2O3 (2-) (aq) + Cl2 (g) + 6H+ (aq) = 2SO4 (-2) (aq) + 2Cl (-) (aq) + 3H2O (l)

Answer 2

The balanced redox reaction in an acidic aqueous solution is
`\( S_2O_3^(2-) + Cl_2 \rightarrow 2SO_4^(2-) + 2Cl^- \)`.

The given reaction is:

`\[ S_2O_3^(2-) \, (aq) + Cl_2 \, (g) = SO_4^(2-) \, (aq) + Cl^- \, (aq) \]`

Here is a step-by-step solution to balance the equation:

• First, assign oxidation states:

`\[ S: +2 \rightarrow +6 \]`

`\[ Cl: 0 \rightarrow -1 \]`

• Half-reactions:

Oxidation:
`\[ S_2O_3^(2-) \rightarrow SO_4^(2-) \]`

Reduction:
`\[ Cl_2 \rightarrow Cl^- \]`

• Balance atoms:

Oxidation:
`\[ S_2O_3^(2-) \rightarrow 2SO_4^(2-) \]`

Reduction:
`\[ Cl_2 \rightarrow 2Cl^- \]`

• Balance charges:

Oxidation:
`\[ S_2O_3^(2-) \rightarrow 2SO_4^(2-) + 10e^- \]`

Reduction:
`\[ Cl_2 + 2e^- \rightarrow 2Cl^- \]`

• Equalize electrons:

Oxidation:
`\[ S_2O_3^(2-) + 10e^- \rightarrow 2SO_4^(2-) + 10e^- \]`

Reduction:
`\[ Cl_2 + 2e^- \rightarrow 2Cl^- \]`

• Finally, add the two half-reactions:

`\[ S_2O_3^(2-) + Cl_2 \rightarrow 2SO_4^(2-) + 2Cl^- \]`

Thus, the balanced redox reaction in an acidic aqueous solution is
`\( S_2O_3^(2-) + Cl_2 \rightarrow 2SO_4^(2-) + 2Cl^- \)`.