Chapter 15, Problem 15.2P

a)

Interpretation Introduction

Interpretation:

The missing cell potentials when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ has to be calculated.

Concept Introduction:

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Where,

${\text{E}}_{\text{cell}}$ =cell potential

${\text{E}}_{\text{+}}$ =cathode potential

${\text{E}}_{\text{-}}$=anode potential

Answer to Problem 15.2P

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.326}\text{V}$ .

i.e.

• $\text{0}\text{.523}\text{V}$versus $\text{S}\text{.H}\text{.E}$= $\text{0}\text{.326}\text{V}$ versus $\text{Ag}|\text{AgCl}$.

Explanation of Solution

Cell Voltage:

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Given,

${\text{E}}_{\text{+}}$= $\text{0}\text{.523}\text{V}$

${\text{E}}_{\text{-}}$= $\text{0}\text{.00}\text{V}$

$\text{0}\text{.523}\text{V}$ Versus $\text{S}\text{.H}\text{.E}$= x Versus $\text{Ag}|\text{AgCl}$.

$\text{R}\text{.H}\text{.S=L}\text{.H}\text{.S}$,

$\text{R}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=}\text{0}\text{.523-}\text{0}\text{.197=0}\text{.326V}$

$\text{L}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=}\text{0}\text{.326V}\text{-0}\text{.00V=0}\text{.326}\text{V}$

So, the value of x is $\text{0}\text{.326}\text{V}$.

Conclusion

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.326}\text{V}$.

b)

Interpretation Introduction

Interpretation:

The missing cell potentials when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ has to be calculated.

Concept Introduction:

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Where,

${\text{E}}_{\text{cell}}$=cell potential

${\text{E}}_{\text{+}}$=cathode potential

${\text{E}}_{\text{-}}$=anode potential

Answer to Problem 15.2P

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.086}\text{V}$ .

i.e.

• $\text{-0}\text{.111}\text{V}$versus $\text{Ag}|\text{AgCl}$= $\text{0}\text{.086}\text{V}$ versus $\text{S}\text{.H}\text{.E}$.

Explanation of Solution

Cell Voltage:

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Given,

${\text{E}}_{\text{+}}$= $\text{-0}\text{.111}\text{V}$

${\text{E}}_{\text{-}}$= $\text{0}\text{.197}\text{V}$

$\text{-0}\text{.111}\text{V}$versus $\text{Ag}|\text{AgCl}$= $\text{x}$ versus $\text{S}\text{.H}\text{.E}$.

$\text{R}\text{.H}\text{.S=L}\text{.H}\text{.S}$,

$\text{R}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=-0}\text{.111+0}\text{.197=0}\text{.086}\text{V}$

$\text{L}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=}\text{0}\text{.086V-0}\text{.00 V=0}\text{.086}\text{V}$

So, the value of x is $\text{0}\text{.086 V}$.

Conclusion

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.086}\text{V}$.

c)

Interpretation Introduction

Interpretation:

The missing cell potentials when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ has to be calculated.

Concept Introduction:

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Where,

${\text{E}}_{\text{cell}}$=cell potential

${\text{E}}_{\text{+}}$=cathode potential

${\text{E}}_{\text{-}}$=anode potential

Answer to Problem 15.2P

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.019 V}$ .

i.e.

• $\text{-0}\text{.222V}$versus $\text{S}\text{.C}\text{.E}$= $\text{x}$ versus $\text{S}\text{.H}\text{.E}$.

Explanation of Solution

Cell Voltage:

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Given,

${\text{E}}_{\text{+}}$= $\text{-0}\text{.222}\text{V}$

${\text{E}}_{\text{-}}$= $\text{+0}\text{.241}\text{V}$

$\text{-0}\text{.222V}$versus $\text{S}\text{.C}\text{.E}$= $\text{x}$ versus $\text{S}\text{.H}\text{.E}$.

$\text{R}\text{.H}\text{.S=L}\text{.H}\text{.S}$,

$\text{R}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=-0}\text{.222+0}\text{.241=0}\text{.019}\text{V}$

$\text{L}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=}\text{0}\text{.019V-0}\text{.00 V=0}\text{.019}\text{V}$

So, the value of x is $\text{0}\text{.019 V}$.

Conclusion

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.019V}$.

d)

Interpretation Introduction

Interpretation:

The missing cell potentials when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ has to be calculated.

Concept Introduction:

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Where,

${\text{E}}_{\text{cell}}$=cell potential

${\text{E}}_{\text{+}}$=cathode potential

${\text{E}}_{\text{-}}$=anode potential

Answer to Problem 15.2P

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{-0}\text{.021}\text{V}$ .

i.e.

• $\text{0}\text{.023V}$versus $\text{Ag}|\text{AgCl}$= $\text{-0}\text{.021}\text{V}$ versus $\text{S}\text{.C}\text{.E}$.

Explanation of Solution

Cell Voltage:

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Given,

${\text{E}}_{\text{+}}$= $\text{0}\text{.023V}$

${\text{E}}_{\text{-}}$= $\text{+0}\text{.197}\text{V}$

$\text{0}\text{.023V}$versus $\text{Ag}|\text{AgCl}$= $\text{x}$ versus $\text{S}\text{.C}\text{.E}$.

$\text{R}\text{.H}\text{.S=L}\text{.H}\text{.S}$,

$\text{R}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=0}\text{.023+0}\text{.197=0}\text{.022}\text{V}$

$\text{L}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=}\text{0}\text{.022V-0}\text{.241V=-0}\text{.021}\text{V}$

So, the value of x is $\text{-0}\text{.021}\text{V}$.

Conclusion

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{-0}\text{.021}\text{V}$.

e)

Interpretation Introduction

Interpretation:

The missing cell potentials when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ has to be calculated.

Concept Introduction:

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Where,

${\text{E}}_{\text{cell}}$=cell potential

${\text{E}}_{\text{+}}$=cathode potential

${\text{E}}_{\text{-}}$=anode potential

Answer to Problem 15.2P

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.021}\text{V}$ .

i.e.

• $\text{-0}\text{.023V}$versus $\text{S}\text{.C}\text{.E}$= $\text{x}$ versus $\text{Ag}|\text{AgCl}$.

Explanation of Solution

Cell Voltage:

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Given,

${\text{E}}_{\text{+}}$= $\text{-0}\text{.023V}$

${\text{E}}_{\text{-}}$= $\text{+0}\text{.241V}$

$\text{-0}\text{.023V}$versus $\text{S}\text{.C}\text{.E}$= $\text{x}$ versus $\text{Ag}|\text{AgCl}$.

$\text{R}\text{.H}\text{.S=L}\text{.H}\text{.S}$,

$\text{R}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=0}\text{.241-0}\text{.023=0}\text{.218}\text{V}$

$\text{L}\text{.H}\text{.S}$,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

${\text{E}}_{\text{cell}}\text{=}\text{0}\text{.218V-0}\text{.197V=0}\text{.021}\text{V}$

So, the value of x is $\text{0}\text{.021}\text{V}$.

Conclusion

The missing cell potential when $\text{Ag}|\text{AgCl}$ and Calomel reference electrodes are saturated with $\text{KCl}$ was calculated as $\text{0}\text{.021}\text{V}$.