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2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)18. Given the balanced ionic equation below: Which half-reaction represents the reduction that occurs?(1) Al → Al³⁺ + 3e¯(2) Al³⁺ + 3e¯ → Al(3) Cu → Cu²⁺+ 2e¯(4) Cu²⁺ + 2e¯ → Cu

Sagot :

INFORMATION:

We have the next balanced ionic equation:

2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)

And we must determine the half-reaction that represents the reduction that occurs

STEP BY STEP EXPLANATION:

To determine it we must use the given information about the ionic charge of atoms given in the formula

We can see that there is a reduction in the formula since we initially have 3Cu²⁺ and then we obtain 3Cu

Since the initial ionic charge for Cu is 2+, we need to add 2e¯ to obtain Cu. Then, the reduction half-reaction would be Cu²⁺ + 2e¯ → Cu

ANSWER:

(4) Cu²⁺ + 2e¯ → Cu

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