Question

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1. (15 points) balance the following redox reaction, using the half - reaction method.

fe²⁺(aq) + cu²⁺(aq) → fe³⁺(aq) + cu(s)

Explanation:

Step1: Identify half - reactions

Oxidation: \(Fe^{2 + }\to Fe^{3 + }\)
Reduction: \(Cu^{2 + }\to Cu\)

Step2: Balance oxidation half - reaction

For \(Fe^{2 + }\to Fe^{3 + }\), the charge is unbalanced. Add an electron to the right side.
\(Fe^{2 + }\to Fe^{3 + }+e^{-}\)

Step3: Balance reduction half - reaction

For \(Cu^{2 + }\to Cu\), add 2 electrons to the left side.
\(Cu^{2 + }+2e^{-}\to Cu\)

Step4: Make electron transfer equal

Multiply the oxidation half - reaction by 2.
\(2Fe^{2 + }\to 2Fe^{3 + }+2e^{-}\)

Step5: Combine half - reactions

\(2Fe^{2 + }+Cu^{2 + }+2e^{-}\to 2Fe^{3 + }+Cu + 2e^{-}\)
Cancel out the electrons on both sides.

Answer:

\(2Fe^{2 + }(aq)+Cu^{2 + }(aq)\to 2Fe^{3 + }(aq)+Cu(s)\)