Chapter 20, Problem 41A

Interpretation Introduction

(a)

Interpretation:

The cell potential is to be calculated.

Concept introduction:

Reduction will occur at that electrode having high reduction potential.

Answer to Problem 41A

$\begin{array}{l}\text{The}\text{given}\text{cell}\text{reaction}\text{is}\\ {\text{2Ag}}^{\text{+2}}\text{+Pb}\rightleftharpoons {\text{Pb}}^{\text{+2}}\text{+2Ag}\\ \text{The}\text{cell}\text{representation}\text{of}\text{given}\text{cell}\text{reaction}\text{is}\\ {\text{Ag}}^{\text{+1}}|\text{Ag}\Vert {\text{Pb}}^{\text{+2}}|\text{Pb}\\ \text{The}\text{cell}\text{potential}\text{is}\text{to}\text{be}\text{calculated}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=0}\text{.7996-}\left(\text{-0}\text{.1262}\right)\\ \text{=0}\text{.9258}\text{V}\end{array}$

The cell potential is $\text{0}\text{.9258}\text{V}$.

Explanation of Solution

Reduction will occur at that electrode having high reduction potential.

$\begin{array}{l}\text{The}\text{given}\text{cell}\text{reaction}\text{is}\\ {\text{2Ag}}^{\text{+2}}\text{+Pb}\rightleftharpoons {\text{Pb}}^{\text{+2}}\text{+2Ag}\\ \text{The}\text{cell}\text{representation}\text{of}\text{given}\text{cell}\text{reaction}\text{is}\\ {\text{Ag}}^{\text{+1}}|\text{Ag}\Vert {\text{Pb}}^{\text{+2}}|\text{Pb}\\ \text{The}\text{cell}\text{potential}\text{is}\text{to}\text{be}\text{calculated}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=0}\text{.7996-}\left(\text{-0}\text{.1262}\right)\\ \text{=0}\text{.9258}\text{V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The cell potential is to be calculated.

Concept introduction:

Reduction will occur at that electrode having high reduction potential

Answer to Problem 41A

$\begin{array}{l}\text{The}\text{given}\text{cell}\text{reaction}\text{is}\\ \text{Mn}\left(\text{s}\right){\text{+Ni}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Mn}}^{\text{+2}}\text{+Ni}\\ \text{The}\text{cell}\text{representation}\text{of}\text{given}\text{cell}\text{reaction}\text{is}\\ \text{Mn}|{\text{Mn}}^{\text{+2}}\Vert {\text{Ni}}^{\text{+2}}|\text{Ni}\\ \text{The}\text{cell}\text{potential}\text{is}\text{to}\text{be}\text{calculated}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.257}\right)\text{-}\left(\text{-1}\text{.185}\right)\\ \text{=0}\text{.928}\text{V}\end{array}$

The cell potential is $\text{0}\text{.928}\text{V}$

Explanation of Solution

Reduction will occur at that electrode having high reduction potential.

$\begin{array}{l}\text{The}\text{given}\text{cell}\text{reaction}\text{is}\\ \text{Mn}\left(\text{s}\right){\text{+Ni}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Mn}}^{\text{+2}}\text{+Ni}\\ \text{The}\text{cell}\text{representation}\text{of}\text{given}\text{cell}\text{reaction}\text{is}\\ \text{Mn}|{\text{Mn}}^{\text{+2}}\Vert {\text{Ni}}^{\text{+2}}|\text{Ni}\\ \text{The}\text{cell}\text{potential}\text{is}\text{to}\text{be}\text{calculated}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.257}\right)\text{-}\left(\text{-1}\text{.185}\right)\\ \text{=0}\text{.928}\text{V}\end{array}$

The cell potential is $\text{0}\text{.928}\text{V}$

(c)

Interpretation Introduction

Interpretation:

The cell potential is to be calculated.

Concept introduction:

Reduction will occur at that electrode having high reduction potential.

Answer to Problem 41A

$\begin{array}{l}\text{The}\text{given}\text{cell}\text{reaction}\text{is}\\ {\text{I}}_{\text{2}}\left(\text{aq}\right)\text{+ Sn}\left(\text{aq}\right)\rightleftharpoons {\text{2I}}^{\text{-}}{\text{+ Sn}}^{\text{+2}}\\ \text{The}\text{cell}\text{representation}\text{of}\text{given}\text{cell}\text{reaction}\text{is}\\ \text{Sn}|{\text{Sn}}^{\text{+2}}\Vert {\text{I}}^{\text{-1}}|{\text{I}}_{\text{2}}\\ \text{The}\text{cell}\text{potential}\text{is}\text{to}\text{be}\text{calculated}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.5355}\right)\text{-}\left(\text{-0}\text{.1375}\right)\\ \text{=0}\text{.6730}\text{V}\end{array}$

The cell potential is $\text{0}\text{.673}\text{V}$

Explanation of Solution

The standard reduction potential of $\text{Sn}|{\text{Sn}}^{\text{+2}}$ is greater than the reduced potential of iodine ion. So, reduction is occurs at $\text{Sn}|{\text{Sn}}^{\text{+2}}$ electrode and oxidation is occurs at ${\text{I}}^{\text{-1}}|{\text{I}}_{\text{2}}$ electrode.

$\begin{array}{l}\text{The}\text{given}\text{cell}\text{reaction}\text{is}\\ {\text{I}}_{\text{2}}\left(\text{aq}\right)\text{+ Sn}\left(\text{aq}\right)\rightleftharpoons {\text{2I}}^{\text{-}}{\text{+ Sn}}^{\text{+2}}\\ \text{The}\text{cell}\text{representation}\text{of}\text{given}\text{cell}\text{reaction}\text{is}\\ \text{Sn}|{\text{Sn}}^{\text{+2}}\Vert {\text{I}}^{\text{-1}}|{\text{I}}_{\text{2}}\\ \text{The}\text{cell}\text{potential}\text{is}\text{to}\text{be}\text{calculated}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.5355}\right)\text{-}\left(\text{-0}\text{.1375}\right)\\ \text{=0}\text{.6730}\text{V}\end{array}$

The cell potential is $\text{0}\text{.673}\text{V}$.