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Difficulty: EasyBalancing Redox Equations and Half-Reactions
Consider the half-reaction representing the oxidation of thiosulfate ions to tetrathionate ions:
2S2O32−(aq)→S4O62−(aq)+ne−2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + n e^-
What is the number of electrons, nn, required to balance the charge in this half-reaction?

Answer: 2

Answer

The number of electrons required to balance the charge in the half-reaction is 2.
To balance a half-reaction, both atom counts and net electric charges must be equal on both sides of the equation. The reactant side contains 2 thiosulfate ions (2S2O32−2\text{S}_2\text{O}_3^{2-}), giving a net charge of 2×(−2)=−42 \times (-2) = -4. The product side contains 1 tetrathionate ion (S4O62−\text{S}_4\text{O}_6^{2-}), giving a net charge of −2-2. Adding 2 electrons (2e−2 e^-) to the product side lowers its total charge to −4-4, equalizing the charge on both sides.

Step-by-Step Solution

1
Calculate the total charge of the reactant species.
Reactant charge = 2 * (-2) = -4.
There are 2 thiosulfate ions, each with an ionic charge of -2.
2
Calculate the net charge of the ionic product species.
Product charge (excluding electrons) = -2.
There is 1 tetrathionate ion with an ionic charge of -2.
3
Equate the overall charges on both sides to solve for the number of electrons n.
-4 = -2 - n, giving n = 2.
Adding 2 electrons (each carrying a -1 charge) to the product side brings the total product charge to -4, matching the reactant side.

Key Concept

Balancing electric charge in oxidation half-reactions
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Topics

Balancing Redox Equations and Half-Reactions

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