Chapter 21, Problem 21.149P

(a)

Interpretation Introduction

Interpretation:

The oxidizing and reducing agents present in given cell reaction 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.

Oxidation: It is the loss of electrons by a species and thus oxidation number is increased.

Reduction: It is the gain of electrons by a species and thus oxidation number is decreased

(b)

Interpretation Introduction

Interpretation:

The $E^{\circ }{}_{cell}$ value for the given reaction has to be calculated.

Concept Introduction:

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

$\begin{array}{l}\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}\\ where,\\ \boxed{\begin{array}{l}\text{n - Number of electrons involved per equivalent of the net redox reaction in the cell}\\ \text{F - Faraday’s Constant }\left(\text{96500 C}\right)\\ {\text{E°}}_{\text{cell}}\text{- Standard cell potential}\text{.}\end{array}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

From the given cell reaction, the balanced reduction reaction has to be written with oxidation number of each element and also the ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ has to be calculated.

Concept Introduction:

Steps in balancing organic redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons.
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.
11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

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

$\begin{array}{l}\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}\\ where,\\ \boxed{\begin{array}{l}\text{n - Number of electrons involved per equivalent of the net redox reaction in the cell}\\ \text{F - Faraday’s Constant }\left(\text{96500 C}\right)\\ {\text{E°}}_{\text{cell}}\text{- Standard cell potential}\text{.}\end{array}}\end{array}$

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

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