Chapter 21.4, Problem 21.8BFP

Interpretation Introduction

Interpretation:

The cell potential for given voltaic cell with two ${\text{Au/Au}}^{\text{3+}}$ half cells has to be calculated. The negative electrode in the given cell has to be identified.

Concept Introduction:

An electrochemical cell is a device in which a redox reaction is used to convert chemical energy into electrical energy. Such device is also known as the galvanic or voltaic cell.

Galvanic cell consists of two half-cells. The redox reaction occurs in these half-cells. The half-cell in which the reduction reaction occurs is known as the reduction half-cell, whereas the half-cell in which the oxidation reaction occurs is known as the oxidation half-cell.

Anode: The electrode where the oxidation occurs is called as an anode. It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode. It is a positively charged electrode.

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}$

n - Number of electrons involved per equivalent of the net redox reaction in the cell

F - Faraday’s Constant (96500 C)

$\text{E}{°}_{\text{cell}}$ - Standard cell potential.

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}$