Chapter 19, Problem 19.93QP

(a)

Interpretation Introduction

Interpretation:

The unknown electrode ‘X’ has to be identified.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The standard reduction potential of Mg (+2) as follows,

$Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V$

The standard reduction potential of Pb (+2) as follows,

$X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}oxidation:Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V\\ \underset{\_}{\mathrm{Re}duction:X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?}\\ \underset{\_}{overall:Mg\left(s\right)+X^{2+}\left(aq\right)\to M{g}^{2+}\left(aq\right)+X\left(s\right)}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}{\text{E}}_{\text{anode}}^{\text{o}}\\ 2.12\text{V}\text{=}\left(\text{?}\right)\text{V}\text{-}\left(\text{-2}\text{.37}\right)\text{V}\\ x\text{=}\text{-0}\text{.25V}\end{array}$

Therefore, the cathode emf is $\text{-0}\text{.25V}$, on checking the table 18.1 of the text, X is NICKEL, Ni

(b)

Interpretation Introduction

Interpretation:

The unknown electrode ‘X’ has to be identified.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The standard reduction potential of Mg (+2) as follows,

$Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V$

The standard reduction potential of Pb (+2) as follows,

$X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}oxidation:Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V\\ \underset{\_}{\mathrm{Re}duction:X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?}\\ \underset{\_}{overall:Mg\left(s\right)+X^{2+}\left(aq\right)\to M{g}^{2+}\left(aq\right)+X\left(s\right)}\end{array}$

The given standard emf of the redox reaction is,$2.24\text{V}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}{\text{E}}_{\text{anode}}^{\text{o}}\\ 2.24\text{V}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}\left(\text{-2}\text{.37}\right)\text{V}\\ {\text{E}}_{\text{cathode}}^{\text{o}}\text{=}\text{-0}\text{.13V}\end{array}$

Therefore, the cathode emf is $\text{-0}\text{.13V}$, on checking the table 18.1 of the text, X is LEAD, Pb

c)

Interpretation Introduction

Interpretation:

The unknown electrode ‘X’ has to be identified.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The standard reduction potential of Mg (+2) as follows,

$Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V$

The standard reduction potential of Pb (+2) as follows,

$X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}oxidation:Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V\\ \underset{\_}{\mathrm{Re}duction:X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?}\\ \underset{\_}{overall:Mg\left(s\right)+X^{2+}\left(aq\right)\to M{g}^{2+}\left(aq\right)+X\left(s\right)}\end{array}$

The given standard emf of the redox reaction is,$1.61\text{V}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}{\text{E}}_{\text{anode}}^{\text{o}}\\ 1.61\text{V}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}\left(\text{-2}\text{.37}\right)\text{V}\\ {\text{E}}_{\text{cathode}}^{\text{o}}\text{=}\text{-0}\text{.76V}\end{array}$

Therefore, the cathode emf is $\text{-0}\text{.76V}$, on checking the table 18.1 of the text, X is ZINC, Zn

d)

Interpretation Introduction

Interpretation:

The unknown electrode ‘X’ has to be identified.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The standard reduction potential of Mg (+2) as follows,

$Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V$

The standard reduction potential of Pb (+2) as follows,

$X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}oxidation:Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}^0{}_{anode}=-2.37V\\ \underset{\_}{\mathrm{Re}duction:X^{2+}\left(aq\right)+2e^{-}\to X\left(s\right)E_{cathode}^{\circ }=?}\\ \underset{\_}{overall:Mg\left(s\right)+X^{2+}\left(aq\right)\to M{g}^{2+}\left(aq\right)+X\left(s\right)}\end{array}$

The given standard emf of the redox reaction is,$1.93\text{V}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}{\text{E}}_{\text{anode}}^{\text{o}}\\ 1.93\text{V}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}\left(\text{-2}\text{.37}\right)\text{V}\\ {\text{E}}_{\text{cathode}}^{\text{o}}\text{=}\text{-0}\text{.44V}\end{array}$

Therefore, the cathode emf is $\text{-0}\text{.44V}$, on checking the table 18.1 of the text, X is IRON, Fe.