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At 298 K, the standard reduction potenti...

At 298 K, the standard reduction potentials for the following half reactions are given. Which acts as anode with others in electrochemical cell
`Zn^(+2)(aq) + 2e^(-) to Zn(s), -0.762`
`Cr^(+3)(aq) + 3e^(-) to Cr(s), -0.740`
`2H^(+)(aq) + 2e^(-) to H_2(g), -0.000`
`Fe^(+3)(aq) + e^(-) to Fe^(2+)(aq) , +0.762`

A

Zn(s)

B

`H_(2)`(g)

C

Cr(s)

D

`Fe^(2+)` (aq)

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The correct Answer is:
To determine which half-reaction acts as the anode in an electrochemical cell, we need to analyze the standard reduction potentials provided for each half-reaction. The anode is the electrode where oxidation occurs, and oxidation potential is the reverse of the reduction potential. ### Step-by-Step Solution: 1. **Identify the Half-Reactions and Their Reduction Potentials**: - \( \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \), \( E^\circ = -0.762 \, \text{V} \) - \( \text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \), \( E^\circ = -0.740 \, \text{V} \) - \( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \), \( E^\circ = 0.000 \, \text{V} \) - \( \text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \), \( E^\circ = +0.762 \, \text{V} \) 2. **Convert Reduction Potentials to Oxidation Potentials**: - For \( \text{Zn} \): - Oxidation: \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \) - Oxidation Potential = \( +0.762 \, \text{V} \) - For \( \text{Cr} \): - Oxidation: \( \text{Cr}(s) \rightarrow \text{Cr}^{3+}(aq) + 3e^- \) - Oxidation Potential = \( +0.740 \, \text{V} \) - For \( \text{H}_2 \): - Oxidation: \( \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \) - Oxidation Potential = \( 0.000 \, \text{V} \) - For \( \text{Fe}^{2+} \): - Oxidation: \( \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^- \) - Oxidation Potential = \( -0.762 \, \text{V} \) 3. **Compare the Oxidation Potentials**: - Oxidation Potentials: - \( \text{Zn} \): \( +0.762 \, \text{V} \) - \( \text{Cr} \): \( +0.740 \, \text{V} \) - \( \text{H}_2 \): \( 0.000 \, \text{V} \) - \( \text{Fe}^{2+} \): \( -0.762 \, \text{V} \) 4. **Determine the Highest Oxidation Potential**: - The highest oxidation potential is \( +0.762 \, \text{V} \) for \( \text{Zn} \). This indicates that zinc has the greatest tendency to be oxidized, making it the anode in the electrochemical cell. 5. **Conclusion**: - Therefore, the half-reaction that acts as the anode is \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \). ### Final Answer: **The half-reaction that acts as the anode is \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \).** ---
To determine which half-reaction acts as the anode in an electrochemical cell, we need to analyze the standard reduction potentials provided for each half-reaction. The anode is the electrode where oxidation occurs, and oxidation potential is the reverse of the reduction potential. ### Step-by-Step Solution: 1. **Identify the Half-Reactions and Their Reduction Potentials**: - \( \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \), \( E^\circ = -0.762 \, \text{V} \) - \( \text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \), \( E^\circ = -0.740 \, \text{V} \) - \( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \), \( E^\circ = 0.000 \, \text{V} \) ...
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