Chapter 21, Problem 21.72P

Interpretation Introduction

Interpretation:

The half-cell that houses the anode has to be identified. The cell voltage has to be calculated.

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

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.

${\text{E}}^{\text{o}}{}_{\text{cell}}$-the cell potential measured at specified temperature conditions which is used to compare the potentials of different cells since, the ${\text{E}}_{\text{cell}}$ is affected by concentration changes as reaction proceeds and due to energy loss from heating of the cell and external circuit.

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

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