Chapter 19, Problem 19.32QP

You have 1.0 M solutions of Al(NO₃)₃ and AgNO₃ along with Al and Ag electrodes to construct a voltaic cell. The salt bridge contains a saturated solution of KCl. Complete the picture associated with this problem by

1. a writing the symbols of the elements and ions in the appropriate areas (both solutions and electrodes).
2. b identifying the anode and cathode.
3. c indicating the direction of electron flow through the external circuit.
4. d indicating the cell potential (assume standard conditions, with no current flowing).
5. e writing the appropriate half-reaction under each of the containers.
6. f indicating the direction of ion flow in the salt bridge.
7. g identifying the species undergoing oxidation and reduction.
8. h writing the balanced overall reaction for the cell.

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

The symbols of elements and ions in cell are,

Symbols of Aluminium is Al, Silver is Ag, Aluminium ion is ${\text{Al}}^{\text{3+}}$, Silver ion is ${\text{Ag}}^{\text{+}}$

Explanation of Solution

The symbols of elements and ions in cell are,

Symbol of Aluminium is Al and it is an anode

Symbol of Silver is Ag and it is an cathode

Symbol of Aluminium ion is ${\text{Al}}^{\text{3+}}$

Symbol of Silver ion is ${\text{Ag}}^{\text{+}}$

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

Silver rod in Silver nitrate is a cathode and Aluminium rod in Aluminium nitrate is an anode.

Explanation of Solution

In given cell, Silver rod in Silver nitrate is a cathode and Aluminium rod in Aluminium nitrate is an anode.

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

$\text{Al}$ is an anode and $\text{Ag}$ is a cathode and electron flow is $\text{Al}\stackrel{}{\to }\text{Ag}$ in direction; therefore cell diagram is drawn in Figure 1.

Figure 1

Explanation of Solution

$\text{Al}$ is an anode and $\text{Ag}$ is a cathode and electron flow is $\text{Al}\stackrel{}{\to }\text{Ag}$ in direction; therefore cell diagram is drawn in Figure 1.

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

The cell potential (EMF) of given voltaic cell is $\text{2}\text{.46}\text{V}$.

Explanation of Solution

The standard reduction potentials of (SRQ) of half cell reactions are record from standard reduction potentials table and they are,

$\begin{array}{l}{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}\text{+ 3e-}\stackrel{}{\to }{\text{Al}}_{\left(\text{s}\right)}{\text{E}}_{\text{red}}\text{ =-1}\text{.66 V}\hfill \\ {\text{Ag}}^{\text{+}}{}_{\text{(aq) }}{\text{+ 2e}}^{\text{-}}\stackrel{}{\to }{\text{ Ag}}_{\left(\text{s}\right)}{\text{E}}_{\text{red}}\text{ = 0}\text{.80 V}\hfill \end{array}$

The most positive SQR is considering as cathode potential.

The SQR of electrodes are plugged in the bellow equation to give cell potential of given voltaic cell.

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cathode}}{\text{-E}}_{\text{anode}}\\ \text{=}\text{0}\text{.80-(-1}\text{.66)}\\ \text{=2}\text{.46}\text{V}\end{array}$

The cell potential (EMF) of given voltaic cell is $\text{2}\text{.46}\text{V}$ so the given cell is spontaneous.

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

The Oxidation half cell reaction is,

${\text{Al}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+ 3e}}^{\text{-}}$

The reduction half cell reaction is,

${\text{3Ag}}^{\text{+}}{}_{\text{(aq) }}{\text{+ 3e}}^{\text{-}}\stackrel{}{\to }{\text{3Ag}}_{\left(\text{s}\right)}$

Explanation of Solution

The Oxidation half cell reaction is,

${\text{Al}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+ 3e}}^{\text{-}}$

The reduction half cell reaction is,

${\text{3Ag}}^{\text{+}}{}_{\text{(aq) }}{\text{+ 3e}}^{\text{-}}\stackrel{}{\to }{\text{3Ag}}_{\left(\text{s}\right)}$

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

${\text{K}}^{\text{+}}$ ion moves to anode half cell side and ${\text{Cl}}^{\text{-}}$ ion moves to cathode half cell side.

Explanation of Solution

Figure 1

${\text{K}}^{\text{+}}$ ion moves to anode half cell side and ${\text{Cl}}^{\text{-}}$ ion moves to cathode half cell side.

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

${\text{Al}}_{\left(\text{s}\right)}$ gets Oxidized and ${\text{Ag}}^{\text{+}}$ gets reduces.

Explanation of Solution

The Oxidation half cell reaction is,

${\text{Al}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+ 3e}}^{\text{-}}$

The reduction half cell reaction is,

${\text{3Ag}}^{\text{+}}{}_{\text{(aq) }}{\text{+ 3e}}^{\text{-}}\stackrel{}{\to }{\text{3Ag}}_{\left(\text{s}\right)}$

Hence, ${\text{Al}}_{\left(\text{s}\right)}$ gets Oxidized and ${\text{Ag}}^{\text{+}}$ gets reduces.

Interpretation Introduction

Interpretation:

Anode, cathode, direction of electron flow, symbols of elements and ions in cell, EMF of the cell and half cell and balanced overall cell reactions should be given.

Concept introduction:

Voltaic cell:

The device, which is converting the chemical energy into electrical energy, is called voltaic cell and this conversation is takes place by the redox reaction.

The oxidation half reaction takes place in anode and reduction half reaction takes place in cathode.

From the result of this redox reaction the electron flow is form anode to cathode direction in outer circuit.

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

The cell potential value is positive in spontaneous cell and negative in nu in spontaneous cell.

Answer to Problem 19.32QP

The balanced overall cell reaction is,

${\text{Al}}_{\left(\text{s}\right)}{\text{+3Ag}}^{\text{+}}{}_{\text{(aq) }}\stackrel{}{\to }{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}+{\text{3Ag}}_{\left(\text{s}\right)}$

Explanation of Solution

The Oxidation half cell reaction is,

${\text{Al}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+ 3e}}^{\text{-}}$

The reduction half cell reaction is,

${\text{3Ag}}^{\text{+}}{}_{\text{(aq) }}{\text{+ 3e}}^{\text{-}}\stackrel{}{\to }{\text{3Ag}}_{\left(\text{s}\right)}$

To sum the two half cell reactions and remove a electron to give a balanced overall cell reaction.

The balanced overall cell reaction is,

${\text{Al}}_{\left(\text{s}\right)}{\text{+3Ag}}^{\text{+}}{}_{\text{(aq) }}\stackrel{}{\to }{\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}+{\text{3Ag}}_{\left(\text{s}\right)}$