Chapter 19, Problem 44E

Interpretation Introduction

(a)

Interpretation:

When the 0.65 M KOH is replaced by 0.65 M NH₃,whether the value of $E_{\text{cell}}$ for 0.65 M NH₃ will be higher or lower than in the cell with 0.65 M KOH should be determined.

Concept introduction:

The relation between hydrogen ion concentration and hydroxide ion concentration is represented as follows:

$\left[{\text{H}}^{+}\right]\times \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

In an electrochemical cell, there are two electrodes that can be observed.

If oxidation takes place on an electrode, that electrode is called anode. The species in that electrode remove electrons and itself gets oxidized.

If reduction takes place on an electrode, that electrode is called cathode. The species in that electrode absorbs electrons and itself gets reduced.

The electrode potential of cell can be calculated as follows:

$E^0{}_{\text{cell}}=E^0{}_{\text{cathode}} -E^0{}_{\text{anode}}$

Nernst equation for a cell can be represented as follows:

${\text{E}}_{cell}={\text{E}}^0{}_{cell}-\frac{0.0592\text{ V}}{\text{z}}\text{ log Q}$

Here,

z = number of moles of electrons transferred in the cell and Q is ratio of concentration of products to reactant in a cell.

Interpretation Introduction

(b)

Interpretation:

The value of $E_{cell}$ should be calculated.

Concept introduction:

Ammonia is weak base. Therefore, partially ionized in water as follows:

${\text{NH}}_{\text{3}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq) + OH}}^{\text{-}}\text{(aq)}$

For the above reaction, the base equilibrium constant can be calculated as follows:

${\text{K}}_b=\frac{\left[{\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq) }\right]\left[{\text{OH}}^{\text{-}}\text{(aq)}\right]}{{\text{NH}}_{\text{3}}\text{(aq) }}$

Here,

$\left[{\text{H}}^{+}\right]\times \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

In an electrochemical cell, there are two electrodes that can be observed.

If oxidation takes place on an electrode, that electrode is called anode. The species in that electrode remove electrons and itself gets oxidized.

If reduction takes place on an electrode, that electrode is called cathode. The species in that electrode absorbs electrons and itself gets reduced.

The electrode potential of cell can be calculated as follows:

$E^0{}_{\text{cell}}=E^0{}_{\text{cathode}} -E^0{}_{\text{anode}}$

Nernst equation for a cell can be represented as follows:

${\text{E}}_{cell}={\text{E}}^0{}_{cell}-\frac{0.0592\text{ V}}{\text{z}}\text{ log Q}$

Here,

z = number of moles of electrons transferred in the cell and Q is ratio of concentration of products to reactant in a cell.