Chapter 17, Problem 17.137MP

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the cell reaction has to be written.

(b)

Interpretation Introduction

Interpretation:

$\text{ΔG°}$ and the equilibrium constant for cell reaction at $\text{25°C}$ 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

Interpretation:

The cell voltage at $\text{25°C}$ when the concentration of $\text{KOH}$ in the electrolyte is $\text{5}\text{.0M}$ has to be given.

(d)

Interpretation Introduction

Interpretation:

The mass of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ formed at the anode when the battery produces constant current at the given time and the number of water molecules consumed in the process has to be given.

Concept Introduction:

Moles of Electrons:

The charge on one mole of electrons is $\text{96,500}\text{C,}$ and the number of moles of electrons passed through the cell is calculated as,

$\text{Moles}\text{of}{\text{e}}^{\text{-}}\text{=}\text{Charge(C)×}\frac{\text{1}\text{mole}{\text{e}}^{\text{-}}}{\text{96,500}\text{C}}$

The flow chart below is useful in the calculation of grams or litres of the product.

Figure 1