Identify the reaction from having top position in EMF series according to their electrode potential at 298k

Options

A. `Mg^(2+) + 2e^- rarr Mg`
B. `Fe^(2+) + 2e^- rarr Fe`
C. `Au^(3+) + 3e^- rarr Au`
D. `K^(+) + 1e^- rarr K`

Answer

To identify the reaction that has the top position in the EMF series according to their electrode potential at 298 K, we will follow these steps:

### Step 1: List the reactions and their standard electrode potentials
We have the following reactions with their respective standard electrode potentials (E°):

1. \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \)
\( E° = -2.37 \, \text{V} \)

2. \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \)
\( E° = -0.44 \, \text{V} \)

3. \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \)
\( E° = +1.50 \, \text{V} \)

4. \( \text{K}^+ + e^- \rightarrow \text{K} \)
\( E° = -2.925 \, \text{V} \)

### Step 2: Compare the standard electrode potentials
To determine which reaction has the highest standard electrode potential, we compare the values:

- \( -2.37 \, \text{V} \) (Mg)
- \( -0.44 \, \text{V} \) (Fe)
- \( +1.50 \, \text{V} \) (Au)
- \( -2.925 \, \text{V} \) (K)

### Step 3: Identify the highest potential
From the comparison, we can see that \( +1.50 \, \text{V} \) (Au) is the highest value among the listed potentials.

### Conclusion
The reaction with the top position in the EMF series is:
\[ \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \]
with a standard electrode potential of \( +1.50 \, \text{V} \).

### Final Answer
The correct option is the third reaction: \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \).

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