Chapter 19, Problem 19.105QP

(a)

Interpretation Introduction

Interpretation:

The standard electrode potential for the given cell has to be calculated in accordance with the given conditions and the test tube which is having ${\text{HNO}}_3$ and $\text{H}g$ has to be identified.

Concept Introduction:

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

  ${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics. 

Free energy and the cell potential is related by the given equation.

  $\text{ΔG=-nFE}$

Where,

  $\text{ΔG}$ is the change in free energy

  $\text{n}$ is the number of electrons transferred

  $\text{F}$ is the Faraday constant $\text{(F}=\text{96485}{\text{.338Cmol}}^{\text{-1}})$

  $\text{E}$ is the cell potential

(b)

Interpretation Introduction

Interpretation:

The standard electrode potential for the given cell has to be calculated in accordance with the given conditions and the test tube which is having ${\text{HNO}}_3$ and $\text{H}g$ has to be identified.

Concept Introduction:

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

  ${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics. 

Free energy and the cell potential is related by the given equation.

  $\text{ΔG=-nFE}$

Where,

  $\text{ΔG}$ is the change in free energy

  $\text{n}$ is the number of electrons transferred

  $\text{F}$ is the Faraday constant $\text{(F}=\text{96485}{\text{.338Cmol}}^{\text{-1}})$

  $\text{E}$ is the cell potential