Chapter 20, Problem 40A

Interpretation Introduction

(a)

Interpretation:

The given reaction is to be balanced.

Concept introduction:

When number of molecules on left side of the reaction is equal to number of molecules on right side of the reaction.

Answer to Problem 40A

The balanced cell reaction I written as

${\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+ Fe}}^{\text{+3}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{Fe}}^{\text{+2}}\left(\text{aq}\right)\text{+}\text{I}{\text{2}}_{}\left(\text{g}\right){\text{+H}}^{\text{+}}{\text{+OH}}^{\text{-}}$

Explanation of Solution

$\begin{array}{l}\text{The}\text{given}\text{represantation}\text{is}\\ {\text{I}}^{\text{-}}|{\text{I}}_{\text{2}}\Vert {\text{Fe}}^{\text{+3}}|{\text{Fe}}^{\text{+2}}\\ \text{so, thereactionis}\\ {\text{I}}^{\text{-}}\left(\text{aq}\right)\text{+}{\text{Fe}}^{\text{+3}}\left(\text{aq}\right)\to {\text{Fe}}^{\text{+2}}\left(\text{aq}\right){\text{+I}}_{\text{2}}\left(\text{g}\right)\\ \text{to}\text{balanced}\text{the}\text{equation, we}\text{multiplying}\text{the}\text{iodide}\text{ion}\text{on}\text{left}\text{side}\text{by}\text{2}\text{and}\text{add}\\ \text{one}\text{water}\text{moleucule}\text{on}\text{left}\text{side}\text{and}\text{one}\text{hydrogen}\text{and}\text{one}\text{hydroxide}\text{ion}\text{on}\text{right}\text{side}\\ \text{atom}\text{on}\text{right}\text{side}\text{by}\text{2}\text{also}\text{.}\end{array}$

Now, the balanced cell reaction is

${\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+ Fe}}^{\text{+3}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{Fe}}^{\text{+2}}\left(\text{aq}\right)\text{+}\text{I}{\text{2}}_{}\left(\text{g}\right){\text{+H}}^{\text{+}}{\text{+OH}}^{\text{-}}$

(b)

Interpretation Introduction

Interpretation:

The given reaction is to be balanced.

Concept introduction:

When number of molecules on left side of the reaction is equal to number of molecules on right side of the reaction.

Answer to Problem 40A

The given reaction is balanced which is given as:

$\text{Sn}\left(\text{s}\right){\text{+2Ag}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Sn}}^{\text{+2}}\left(\text{aq}\right)\text{+2Ag}\left(\text{s}\right)$

Explanation of Solution

$\begin{array}{l}\text{The}\text{given}\text{represantation}\text{is}\\ \text{Sn}|{\text{Sn}}^{\text{+2}}\Vert {\text{Ag}}^{\text{+1}}|\text{Ag}\\ \text{so, thereactionis}\\ \text{Sn}\left(\text{s}\right){\text{+Ag}}^{\text{+}}\left(\text{aq}\right)\to {\text{Sn}}^{\text{+2}}\left(\text{aq}\right)\text{+Ag}\left(\text{s}\right)\\ \text{to}\text{balanced}\text{the}\text{equation, we}\text{multiplying}\text{the}\text{silver}\text{ion}\text{on}\text{left}\text{side}\text{by}\text{2}\text{and}\text{silver}\\ \text{atom}\text{on}\text{right}\text{side}\text{by}\text{2}\text{also}\text{.}\\ \text{Now,}\text{the}\text{balanced}\text{reaction}\text{is}\text{written}\text{as}\\ \text{Sn}\left(\text{s}\right){\text{+2Ag}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Sn}}^{\text{+2}}\left(\text{aq}\right)\text{+2Ag}\left(\text{s}\right)\end{array}$

(c)

Interpretation Introduction

Interpretation:

The given reaction is to be balanced.

Concept introduction:

When number of molecules on left side of the reaction is equal to number of molecules on right side of the reaction.

Answer to Problem 40A

$\begin{array}{l}\text{The}\text{balanced}\text{reaction}\text{is}\text{written}\text{as}\\ \text{Zn}\left(\text{s}\right){\text{+Cd}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{+2}}\left(\text{aq}\right)\text{+Cd}\left(\text{s}\right)\end{array}$

Explanation of Solution

$\begin{array}{l}\text{The}\text{given}\text{represantation}\text{is}\\ \text{Zn}|{\text{Zn}}^{\text{+2}}\Vert {\text{Cd}}^{\text{+2}}|\text{Cd}\\ \text{so, thereactionis}\\ \text{Zn}\left(\text{s}\right)\text{+}{\text{Cd}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{+2}}\left(\text{aq}\right)\text{+Cd}\left(\text{s}\right)\\ \text{the}\text{given}\text{equation}\text{is}\text{balanced}\text{beacuse}\text{the}\text{oxidation}\text{number}\text{othe}\text{balanced}\text{reaction}\text{is}\text{written}\text{as}\\ \text{Zn}\left(\text{s}\right){\text{+Cd}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{+2}}\left(\text{aq}\right)\text{+Cd}\left(\text{s}\right)\end{array}$