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Chapter 17: Problem 47

Write the balanced half-reaction for the conversion of \(\mathrm{Fe}(s)\) to \(\mathrm{Fe}^{2+}(a q) .\)

Short Answer

Expert verified
The balanced half-reaction for the conversion of \(\mathrm{Fe}(s)\) to \(\mathrm{Fe}^{2+}(a q)\) is \(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(a q) + 2e^-.\)

Step by step solution

01

Write down the unbalanced reaction

Start by writing down the unbalanced reaction: \(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(a q)\).
02

Balance the atoms first

Balance the atoms that are not oxygen or hydrogen first. Iron (\(Fe\)) is already balanced in this reaction: \[\(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(a q)\]
03

Balance the charges

Balance the charges by adding electrons (\(e^-\)) to the appropriate side. Since the \(Fe^{2+}\) ion has a charge of +2, add 2 electrons to the right side to balance the charges and produce a net charge of zero in the reaction. \[\(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(a q) + 2e^-\]

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Most popular questions from this chapter

What name is given to an electrode at which oxidation occurs? at which reduction occurs?

Using half-reactions, balance the redox equation of \(\mathrm{Zn}(s)\) and \(\mathrm{Fe}^{3+}(a q)\) reacting to form \(\mathrm{Zn}^{2+}(a q)\) and \(\mathrm{Fe}^{2+}(a q) .\)

Using half-reactions, balance the equation for the redox reaction when \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) and \(\mathrm{Fe}^{2+}(a q)\) react to form \(\mathrm{Cr}^{3+}(a q)\) and \(\mathrm{Fe}^{3+}(a q)\) in acidic solution.

How does the oxidation number of an atom of Mn change in the reaction below? $$\begin{array}{r}{4 \mathrm{Mn}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+} \\ {15 \mathrm{H}_{2} \mathrm{P}_{4} \mathrm{O}_{7}^{2-}(a q) \rightarrow 12 \mathrm{H}_{2} \mathrm{O}(l)+} \\\ {\quad 5 \mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{P}_{4} \mathrm{O}_{7}\right)_{3}^{3-}(a q)}\end{array}$$

Why does a sacrificial anode provide cathodic protection?
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