Chapter 18, Problem 18.158MP

Interpretation Introduction

Interpretation:

The electrode potential of an unknown half-cell can be determined form the analogous free energy value.

Concept introduction:

• Free energy and cell potential: The chemical reaction occurring in a galvanic cell is exothermic and if carried out directly a quantity of heat will be involved. The same change processed through a voltaic cell yields electrical energy. If E be the cell potential of the cell supplying a quantity of electricity Q, then electrical change is EQ. With each equivalent of reactant involved is associated one Faraday (F) of electricity. If n equivalents of reaction occurs, then electrical energy produced = $nFE$. The decrease of the free energy of the process is equal to the electrical energy obtained from a galvanic cell. So $\Delta G=-nFE$.
• Cell Voltage/standard cell potential: In a cell the electron flow from anode to cathode but the electric flow from cathode to anode, thus the cell potential which is also called the electromotive force (e.m.f.) of a cell is the difference between the half-cell potential at cathode and anode. Thus:

  $E_{Cell}=E_{Cathode}-E_{Anode}$
• Cell potential with equilibrium constant and formation constant: The equilibrium constant and formation constant of a reaction can be determined from the cell potential by the use of Nernst equation: Let assume a reaction:

  $aA+bB\rightleftarrows cC+dD$.
Thus the cell potential of the reaction from Nernst equation is:

  $E_{Cell}=E_{Cell}^0-\frac{RT}{nF}\times \ln \frac{{[C]}^c\times {[D]}^d}{{[A]}^a\times {[B]}^b}$

Now at equilibrium the $E_{Cell}$ =0 and the $\frac{{[C]}^c\times {[D]}^d}{{[A]}^a\times {[B]}^b}$ = Equilibrium constant, K.
Thus $\ln K=\frac{nF{E}^0}{RT}$.
On the other hand the formation constant $K_f$ can be determined at any point of the reaction but putting the value of $E_{Cell}$ at that moment.

Given:

The $\Delta G^0$ of the reaction $B{a}^{2+}(aq)+2C{l}^{-}(aq)\to BaC{l}_2(s)$ is 16.7kJ.

To determine:

The standard electrode potential of the reaction $B{a}^{2+}(aq)+2e^{-}\to Ba(s)$.