Find the number of electrons transferred in the equation `Cu(g) + 2Ag^+(aq) → Cu^(2+)(aq) + 2Ag(s)`.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of electrons transferred in the reaction:

\[ \text{Cu(g)} + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s) \]

we can analyze the oxidation and reduction processes occurring in the reaction.

### Step 1: Identify Oxidation and Reduction
- **Oxidation**: In this reaction, copper (Cu) is oxidized. It starts with an oxidation state of 0 in the gaseous state and ends with an oxidation state of +2 in the aqueous state (\( \text{Cu}^{2+} \)). This means that copper loses 2 electrons.

\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]

- **Reduction**: Silver ions (\( \text{Ag}^+ \)) are reduced. Each silver ion gains one electron to form solid silver (Ag). Since there are 2 silver ions, a total of 2 electrons are gained.

\[ 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \]

### Step 2: Combine the Half-Reactions
Now we can combine the half-reactions to see the overall electron transfer:

- The oxidation half-reaction shows that 2 electrons are released by copper.
- The reduction half-reaction shows that 2 electrons are consumed by the silver ions.

### Step 3: Determine Total Electrons Transferred
Since both half-reactions involve 2 electrons, the total number of electrons transferred in the reaction is:

\[ \text{Total electrons transferred} = 2 \]

### Conclusion
Thus, the number of electrons transferred in the reaction is **2**.

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