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Balance the following redox reaction in acidic solution. i2(s) +zn2 (aq)→ io−3(aq) +zn(s)

User Kani
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Final answer:

To balance the given redox reaction in acidic solution, follow these steps: separate the reaction into half-reactions, balance the electrons in each half-reaction, add the balanced half-reactions together, and simplify the equation. The balanced equation for the reaction is 3Zn(s) + I₂(s) + 3H₂O(l) -> 3Zn²+(aq) + 6H⁺(aq) + IO₃¯(aq).

Step-by-step explanation:

To balance the given redox reaction in acidic solution, we need to follow the step-by-step process. First, separate the reaction into two half-reactions, one for oxidation and one for reduction. The oxidation half-reaction involves the formation of zinc ions from solid zinc:

Zn(s) -> Zn²+(aq) + 2e¯

The reduction half-reaction involves the formation of iodate ions from iodine:

I₂(s) + 6H₂O(l) + 6e¯ -> 2IO₃¯(aq) + 12H⁺(aq)

Next, balance the number of electrons in both half-reactions by multiplying them so that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. In this case, the oxidation half-reaction needs to be multiplied by 6:

6Zn(s) -> 6Zn²+(aq) + 12e¯

Now, add the two balanced half-reactions together, canceling out the electrons:

6Zn(s) + I₂(s) + 6H₂O(l) -> 6Zn²+(aq) + 12e¯ + 2IO₃¯(aq) + 12H⁺(aq)

Finally, simplify the equation by canceling out common species:

3Zn(s) + I₂(s) + 3H₂O(l) -> 3Zn²+(aq) + 6H⁺(aq) + IO₃¯(aq)

User Liza Daly
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