Chapter 17, Problem 17.91SP

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the overall cell reaction has to be given.

Concept Introduction:

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are known, the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The change in free energy and the equilibrium constant has to be calculated.

Concept Introduction:

Standard cell potential and equilibrium constant:

The standard cell potential related to the free-energy is given by the equation,

$\text{ΔG°}\text{=}\text{-nFE°}$

The free-energy change is also related to equilibrium constant by the equation,

$\text{ΔG°}\text{=}\text{-RTlnK}$

Combine both the equations,

$\begin{array}{l}\text{-nFE°}\text{=}\text{-RTlnK}\\ \\ \text{E°}\text{=}\frac{\text{RT}}{\text{nFlnK}}\text{=}\frac{\text{2}\text{.303RT}}{\text{nF}}\text{logK}\\ \\ \text{(or)}\\ \\ \text{E°}\text{=}\frac{\text{0}\text{.0592V}}{\text{n}}\text{logK}\end{array}$

Where,

$\begin{array}{l}\text{n}\text{=}\text{number}\text{of}\text{moles}\text{of}\text{electrons}\text{transferred}\\ \text{F}\text{=}\text{Faraday}\text{constant}\text{(96,500}\text{C/mole}{\text{e}}^{\text{-}}\text{)}\\ \text{E°}\text{=}\text{Standard cell potential}\\ \text{ΔG°}\text{=}\text{Change in free-energy}\\ \text{T}\text{=}\text{Temperature in kelvin}\\ \text{R}\text{=}\text{Universal gas constant}\\ \text{K}\text{=}\text{Equilibrium constant}\end{array}$

(c)

Interpretation Introduction

The effect on the cell voltage of a tenfold change in the concentration of $\text{KOH}$ in the electrolyte has to be given and explained