Chapter 14, Problem 14.12P

(a)

Interpretation Introduction

Interpretation:

Cell potential for each half reaction has to be found and net cell voltage has to be calculated.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute $E_{\text{cell}}$ as per the Nernst equation is written as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{n}\log \frac{\left[P\right]}{\left[R\right]}$

Here,

$E_{\text{cell}}$ denotes overall cell potential.

$E_{\text{cell}}^{°}$ denotes standard cell potential.

$n$ is moles of electrons transferred in each reaction.

$\left[P\right]$ denotes concentration of products.

$\left[R\right]$ denotes concentration of reactant.

(b)

Interpretation Introduction

Interpretation:

Cell-diagram has to be drawn.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

Net cell potential for overall reaction has to be calculated.

Concept Introduction:

Cell potential is evaluated by expression as follows:

  $E_{\text{cell}}^{°}=E_{\text{oxidation}}^{°}+E_{\text{reduction}}^{°}$