Balance the following redox reaction using the half reaction method;
S2O3 2-(aq) + I2(aq)—-> I-(aq) + S4O6 2- (aq)
- Write out and balance the oxidation protion of this reaction.
- Write out and balance the reduction portion of this reaction.
- What is the overall reaction for this reaction?
The Correct Answer and Explanation is :
Step 1: Write and Balance the Oxidation Half-Reaction
Oxidation occurs when a species loses electrons. Here, thiosulfate (( S_2O_3^{2-} )) is oxidized to tetrathionate (( S_4O_6^{2-} )).
- Identify oxidation change:
[
2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^-
]
Sulfur goes from oxidation state (+2) in ( S_2O_3^{2-} ) to a mixed oxidation state in ( S_4O_6^{2-} ). - Charge balance:
- Left side: ( 2(-2) = -4 )
- Right side: ( -2 )
- To balance charge, add ( 2e^- ) on the right.
Final oxidation half-reaction:
[
2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^-
]
Step 2: Write and Balance the Reduction Half-Reaction
Reduction occurs when a species gains electrons. Here, iodine (( I_2 )) is reduced to iodide (( I^- )).
- Identify reduction change:
[
I_2 + 2e^- \rightarrow 2I^-
]
Iodine changes from oxidation state ( 0 ) in ( I_2 ) to ( -1 ) in ( I^- ).
Final reduction half-reaction:
[
I_2 + 2e^- \rightarrow 2I^-
]
Step 3: Overall Balanced Reaction
To obtain the overall balanced redox reaction, add both half-reactions:
[
2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^-
]
[
I_2 + 2e^- \rightarrow 2I^-
]
Cancel electrons:
[
2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-
]
This is the final balanced equation.
Explanation (300 Words)
Redox (reduction-oxidation) reactions involve the transfer of electrons between reactants. In this reaction, thiosulfate (( S_2O_3^{2-} )) is oxidized to tetrathionate (( S_4O_6^{2-} )), while iodine (( I_2 )) is reduced to iodide (( I^- )). To balance this reaction using the half-reaction method, we split it into oxidation and reduction parts.
First, in the oxidation half-reaction, thiosulfate loses electrons to form tetrathionate. Since the oxidation state of sulfur changes, two thiosulfate ions are needed to generate one tetrathionate ion while releasing two electrons. In the reduction half-reaction, iodine gains two electrons to form two iodide ions.
By ensuring both half-reactions have the same number of electrons, we add them together to form the overall balanced equation:
[
2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-
]
This reaction is important in analytical chemistry, particularly in iodometric titrations, where thiosulfate is used to determine the concentration of iodine. The reaction is also widely used in redox studies because of its clear electron transfer.
Below is an image representation of this reaction.
Here is the chemistry illustration depicting the redox reaction between thiosulfate (( S_2O_3^{2-} )) and iodine (( I_2 )). The image visually represents the oxidation and reduction half-reactions with molecular diagrams and electron transfers. Let me know if you need any modifications!
