Chapter 19, Problem 19.160MP

Interpretation Introduction

(a)

Interpretation:

A balanced equation for the cell reaction should be written.

Concept introduction:

To obtain the cell reaction, cathode reaction and the anode reaction should be added together and before that, the two half-cell reactions should be electronically equal, so that the electrons are cancelled off when adding. If common ions present they should also be deducted.

Interpretation Introduction

(b)

Interpretation:

$\Delta G^0$ and equilibrium concentration K for the cell reaction at 25⁰ C should be calculated.

Concept introduction:

The standard Gibbs free energy change can be calculated as follows:

$\begin{array}{l}\Delta G^0=-nF{E}^0\\ \\ \Delta G^0=\text{ Standard Gibbs free energy}\\ \text{n = number of moles of electrons transferred}\\ \text{F = Faraday constant}\\ {\text{E}}^0=\text{ Standard cell potential}\end{array}$

It is related to equilibrium constant as follows:

$\begin{array}{l}\Delta G^0=-RT\ln K\\ \\ \Delta G^0=\text{ Standard Gibbs free energy}\\ \text{R = Universal gas constant}\\ \text{T = Absolute temperature}\\ \text{K = Equilibrium constant}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The cell voltage at 25⁰ C when the concentration of $\text{KOH}$ in the electrolyte is $\text{5}\text{.0 M}$ should be calculated.

Concept introduction:

The Nernst equation allows to calculate cell potential at non-standard conditions.

$E=E^0-\frac{0.0592\text{ V}}{n}\log Q$

E − non-standard cell potential

E⁰ − standard cell potential

n − number of electrons passed through the cell

Q − reaction quotient

Interpretation Introduction

(d)

Interpretation:

Number of grams of ${\text{Fe(OH)}}_2$ are formed at the anode when the battery produces a constant current of $0.250\text{ A for 40}\text{.0 min}$ and number of water molecules consumed in the process should be calculated.

Concept introduction:

The number of coulombs of charge passed through the cell equals the product of the current in amperes and the time in seconds.

$\text{Charge (C) = Current (A) }\times \text{ Time (s)}$