Question

Interactive Activity—The Relationship among E°cell, Keq, and Gibbs Free Energy In an electrochemical cell, the potential difference between two electrodes under standard conditions is known as the standard cell potential (E∘cell). The standard cell potential can be used to identify the overall tendency of a redox reaction to occur spontaneously. The spontaneity of a reaction is identified using the Gibbs free energy ΔG∘. ΔG∘ is related to E∘cell. E∘cell and ΔG∘ are also related to equilibrium constant Keq of the reaction. Select the image to explore the activity that shows how E∘cell, Keq, and ΔG∘ are related to each other. launch activity In the activity, you should see a triangle, whose three vertices represent Keq, E∘cell, and ΔG∘. You can select two vertices and determine the relation between them. You can then reset the activity and select the next two quantities. Constants The following values may be useful when solving this tutorial. Constant Value E∘Cu 0.337 V E∘Ni -0.257 V R 8.314 J⋅mol−1⋅K−1 F 96,485 C/mol T 298 K Part A In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2+(aq)+2e−→Cu(s) and Ni(s)→Ni2+(aq)+2e− The net reaction is Cu2+(aq)+Ni(s)→Cu(s)+Ni2+(aq) Use the given standard reduction potentials in your calculation as appropriate. Express your answer numerically to three significant figures.

Answer 1

Answer:

Keq = 1.17 × 10²⁰

Explanation:

Let's consider the following redox reaction.

Cu²⁺(aq) + Ni(s) → Cu(s) + Ni²⁺(aq)

We can identify 2 half-reactions.

Cathode (reduction): Cu²⁺(aq) + 2 e⁻ → Cu(s)            E°red = 0.337 V

Anode (oxidation): Ni(s) → Ni²⁺(aq) + 2 e⁻                  E°red = -0.257 V

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.337 V - (-0.257V) = 0.594 V

We can calculate the equilibrium constant (Keq) using the following expression.

$E\°=\frac{0.0592V}{n} logKeq$

where,

n are the moles of electrons transferred

$0.594V=\frac{0.0592V}{2} logKeq\\Keq = 1.17 \times 10^{20}$