Chapter U5, Problem C21.5RE

Interpretation Introduction

(a)

Interpretation:

The metal that will be oxidized and the metal that will be reduced when an electrochemical cell is formed with lead nitrate and silver nitrate must be explained. Balanced half equations for these reactions must also be written.

Concept introduction:

Metal will be oxidized for which reduction potential is less. It will be a reducing agent.

Answer to Problem C21.5RE

Lead metal will be oxidized and silver metal will be reduced.

$\begin{array}{l}\text{Pb}\to {\text{Pb}}^{\text{2+}}\text{+2e}\text{.}\\ {\text{2Ag}}^{\text{+}}\text{+2e}\to \text{2Ag}\text{.}\end{array}$

Explanation of Solution

As the silver electrode has a higher reduction potential as compared to the lead electrode, silver will be cathode where reduction will take place. Thus silver will be reduced and lead will be oxidized.

The two half cell reactions are shown in which 2e are involved.

Interpretation Introduction

(b)

Interpretation:

The balanced overall reaction must be written for the cell.

Concept introduction:

Overall electrochemical reaction is the total reaction taking place in cathode and anode which includes oxidation of one electrode and reduction of another electrode.

Answer to Problem C21.5RE

The overall cell reaction is ${\text{Pb(s)+2Ag}}^{\text{+}}\text{(aq)}\to {\text{Pb}}^{\text{2+}}\text{(aq)+2Ag(s)}$

Explanation of Solution

The overall reaction for the cell is written in which electrons are cancelled from both the sides.

The physical states of all the species involved are given in bracket. ${\text{Ag}}^{+}$ and ${\text{Pb}}^{2+}$ will be in aqueous solution and metallic Ag and Pb will be in solid-state.

Interpretation Introduction

(c)

Interpretation:

The number moles of electrons that will flow from cathode to anode must be calculated.

Concept introduction:

Silver is monovalent cation whereas lead is divalent cation. So, 2 moles of silver ions are reacting with 1 mole of lead.

Answer to Problem C21.5RE

2 moles of electron will flow from anode to cathode.

Explanation of Solution

$\begin{array}{l}\text{Pb}\to {\text{Pb}}^{\text{2+}}\text{+2e}\text{.}\\ {\text{2Ag}}^{\text{+}}\text{+2e}\to \text{2Ag}\text{.}\end{array}$

From the two half cell reactions, it can be seen that 2 moles of electrons are involved.

So from anode to cathode, 2 moles of electrons will flow.

Interpretation Introduction

(d)

Interpretation:

The voltage for the electrochemical cell must be calculated.

Concept introduction:

The voltage of an electrochemical cell can be written as

${\text{E}}^{\text{0}}{}_{\text{cell}}{\text{=E}}^{\text{0}}{}_{\text{cathode}}{\text{-E}}^{\text{0}}{}_{\text{anode}}$ where ${\text{E}}^{\text{0}}{}_{\text{cathode}}{\text{ and E}}^{\text{0}}{}_{\text{anode}}$ are standard reduction potentials of cathode and anode respectively.

Answer to Problem C21.5RE

Voltage of the electrochemical cell is 0.93 V.

Explanation of Solution

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

The standard reduction potential for silver (cathode) and lead (anode) electrode are 0.80 V and -0.13 V respectively.

Thus, ${\text{E}}^{\text{0}}{}_{\text{cell}}\text{=[0}\text{.80-(-0}\text{.13)] V=[0}\text{.80+0}\text{.13] V=0}\text{.93 V}\text{.}$