QuestionJune 7, 2025

What is the cell potential of an electrochemical cell that has the half-reactions shown below? Ag^++e^-arrow Ag Fearrow Fe^3++3e^- Click for a reduction potential chart A. -0.44 B. 1.24 V C. 0.44V D. -1.24V

What is the cell potential of an electrochemical cell that has the half-reactions shown below? Ag^++e^-arrow Ag Fearrow Fe^3++3e^- Click for a reduction potential chart A. -0.44 B. 1.24 V C. 0.44V D. -1.24V
What is the cell potential of an electrochemical cell that has the half-reactions
shown below?
Ag^++e^-arrow Ag
Fearrow Fe^3++3e^-
Click for a reduction potential chart
A. -0.44
B. 1.24 V
C. 0.44V
D. -1.24V

Solution
4.4(235 votes)

Answer

0.76 V Explanation 1. Identify Reduction Potentials Ag⁺ + e⁻ → Ag has a reduction potential of +0.80 V; Fe³⁺ + 3e⁻ → Fe has a reduction potential of -0.04 V. 2. Determine Oxidation Reaction Reverse the Fe reaction: Fe → Fe³⁺ + 3e⁻, oxidation potential is +0.04 V. 3. Calculate Cell Potential **E_{cell} = E_{cathode} - E_{anode}**. Cathode (Ag) is +0.80 V, Anode (Fe) is +0.04 V. 4. Compute Final Value E_{cell} = 0.80 V - 0.04 V = 0.76 V

Explanation

1. Identify Reduction Potentials<br /> Ag⁺ + e⁻ → Ag has a reduction potential of +0.80 V; Fe³⁺ + 3e⁻ → Fe has a reduction potential of -0.04 V.<br /><br />2. Determine Oxidation Reaction<br /> Reverse the Fe reaction: Fe → Fe³⁺ + 3e⁻, oxidation potential is +0.04 V.<br /><br />3. Calculate Cell Potential<br /> **E_{cell} = E_{cathode} - E_{anode}**. Cathode (Ag) is +0.80 V, Anode (Fe) is +0.04 V. <br /><br />4. Compute Final Value<br /> E_{cell} = 0.80 V - 0.04 V = 0.76 V
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