Q45P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance each skeleton reaction, calculate ${E^{\circ }}_{cell}$ and state whether the reaction is spontaneous:

(a) ${\text{Cu}}^{\text{ + }}{\text{(aq) + PbO}}_{\text{2}}\text{(s) + S}{{\text{O}}_{\text{4}}}^{\text{2 - }}{\text{(aq)PbSO}}_{\text{4}}{\text{(s) + Cu}}^{\text{2 + }}\text{(aq)[acidic]}$

(b) ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{(aq) + Ni}}^{\text{2 + }}{\text{(aq)O}}_{\text{2}}\text{(g) + Ni(s)[acidic]}$

(c) ${\text{MnO}}_{\text{2}}{\text{(s) + Ag}}^{\text{ + }}\text{(aq)Mn}{{\text{O}}_{\text{4}}}^{\text{ - }}\text{(aq) + Ag(s)[basic]}$

Step-by-Step Solution

Answer

The required work is used to acidic and basic solution also used to get the standard reduction and oxidation divide into half reaction method of spontaneous and non-spontaneous 

Step 1: Defind the standard oxidation and reduction equations of redox reaction

(a) To balance each reaction, divide into half reactions and balance the redox through the half-reaction method

${\text{Cu}}_{\left(aq\right)}^{+}+{\text{PbO}}_{2\left(s\right)}+{\text{SO}}_{4\left(aq\right)}^{2-}\to {\text{PbSO}}_{4\left(s\right)}+{\text{Cu}}_{\left(aq\right)}^{2+}\left[\text{ acidic }\right]$

Separate the half reactions.

${\text{OHR:Cu}}^{\text{ + }}\to {\text{Cu}}^{\text{2 + }}{\text{RHR:PbO}}_{\text{2}}{\text{ + SO}}_{\text{4}}^{\text{2 - }}\to {\text{PbSO}}_{\text{4}}$

Balance electrons and non- $\text{O}$ atoms.

${\text{OHR:Cu}}^{\text{ + }}\to {\text{Cu}}^{\text{2 + }}{\text{ + e}}^{\text{ - }}$

$\text{RHR }:{\text{PbO}}_2+{\text{SO}}_4^{2-}+2e^{-}\to {\text{PbSO}}_4$

Balance reaction charges with  ${\text{H}}^{\text{ + }}$ (acidic environment).

${\text{OHR:Cu}}^{\text{ + }}\to {\text{Cu}}^{\text{2 + }}{\text{ + e}}^{\text{ - }}$

${\text{RHR:PbO}}_{\text{2}}{\text{ + SO}}_{\text{4}}^{\text{2 - }}{\text{ + 2e}}^{\text{ - }}{\text{ + 4H}}^{\text{ + }}\to {\text{PbSO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$width="395" style="max-width: none; vertical-align: -11px;" ${\text{RHR:PbO}}_{\text{2}}{\text{ + SO}}_{\text{4}}^{\text{2 - }}{\text{ + 2e}}^{\text{ - }}{\text{ + 4H}}^{\text{ + }}\to {\text{PbSO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$                                                                 Balance atoms $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ to the opposite side.

${\text{OHR:Cu}}^{\text{ + }}\to {\text{Cu}}^{\text{2 + }}{\text{ + e}}^{\text{ - }}$

${\text{RHR:PbO}}_{\text{2}}{\text{ + SO}}_{\text{4}}^{\text{2 - }}{\text{ + 2e}}^{\text{ - }}{\text{ + 4H}}^{\text{ + }}\to {\text{PbSO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$

Step 2: Finding multiply reactions by least common factor and combine half reactions

$2xOHR:2C{u}^{+}\to 2C{u}^{2+}+2e^{-}\text{RHR }:Pb{O}_2+S{O}_4^{2-}+2e^{-}+4H^{+}\to PbS{O}_4+2H_{2}O\text{ Reaction }:Pb{O}_{2\left(s\right)}+S{O}_{4\left(aq\right)}^{2-}+2C{u}_{\left(aq\right)}^{+}+4H_{\left(aq\right)}^{+}\to PbS{O}_{4\left(s\right)}+2H_2{O}_{\left(l\right)}+2C{u}_{\left(aq\right)}^{2+}$

${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Cu}}^{\text{ + }}}^{\text{o}}\text{ = 0.15V}$

${\text{E}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{PbO}}_{\text{2}}}^{\text{o}}\text{ = 1.70V}$

${\text{E}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}_{\text{oxidation }}^{\text{o}}$

${\text{ = E}}_{{\text{PbO}}_{\text{2}}}^{\text{o}}{\text{ - E}}_{{\text{Cu}}^{\text{ + }}}^{\text{o}}$

$=1.70-0.15=1.55Vcell=1.55V,$spontaneous reaction

Step 3: Given half-reaction acidic method

To balance each reaction, divide into half reactions and balance the redox through the half-reaction method [acidic].

${\text{H}}_2{\text{O}}_{2\left(aq\right)}+\text{Ni}{i}_{\left(aq\right)}^{2+}\to {\text{O}}_{2\left(g\right)}+\text{Ni}{i}_{\left(s\right)}\left[\text{ acidic }\right]$

Separate the half reactions.

${\text{OHR:H}}_{\text{2}}{\text{O}}_{\text{2}}\to {\text{O}}_{\text{2}}{\text{RHR:Nii}}^{\text{2 + }}\to \text{Ni}$

 Multiply reactions by least common factor and combine half reactions, and cancel common ions/ compounds both sides accordingly.

$\text{OHR}:{\text{H}}_2{\text{O}}_2\to {\text{O}}_2+2e^{-}+2{\text{H}}^{+}$

$\text{RHR}:{\text{Ni}}^{2+}+2e^{-}\to \text{Ni Reaction }:{\text{H}}_2{\text{O}}_{2\left(aq\right)}+\text{Ni}{i}_{\left(aq\right)}^{2+}\to 2{\text{H}}_{\left(aq\right)}^{+}+N{i}_{\left(s\right)}+{\text{H}}_2{\text{O}}_{2\left(aq\right)}$

${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{O}}_{\text{2}}}^{\text{o}}\text{ = 0.68V}$

${\text{E}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{Ni}}^{\text{2 + }}}^{\text{o}}\text{ = - 0.25V}$

${\text{E}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}^{\text{o}}\text{ oxidation }$

${\text{ = E}}_{{\text{Ni}}^{\text{2 + }}}^{\text{o}}{\text{ - E}}_{{\text{O}}_{\text{2}}}^{\text{o}}$

$\text{ = - 0.25 - 0.68 = - 0.93V}$

${\text{E}}_{\text{cell }}^{\text{o}}\text{ = - 0.93V}$, non-spontaneous reaction

Step 4: Given half-reaction basic method

To balance each reaction, divide into half reactions and balance the redox through the half-reaction method [basic].

${\text{MnO}}_{2\left(s\right)}+{\text{Ag}}_{\left(aq\right)}^{+}\to {\text{MnO}}_{\text{4(aq)}}^{-}+{\text{Ag}}_{\left(s\right)}\left[basic\right]$

Separate the half reactions.

${\text{OHR:MnO}}_{\text{2}}\to {\text{MnO}}_{\text{4}}^{\text{ - }}$

${\text{RHR:Ag}}^{\text{ + }}\to \text{Ag}$

Balance electrons and non-$\text{O}$ atoms.

${\text{OHR:MnO}}_{\text{2}}\to {\text{MnO}}_{\text{4}}^{\text{ - }}{\text{ + 3e}}^{\text{ - }}{\text{RHR:Ag}}^{\text{ + }}{\text{ + e}}^{\text{ - }}\to \text{Ag}$

 Balance reaction charges with ${\text{OH}}^{\text{ - }}$ (basic environment).

${\text{OHR:MnO}}_{\text{2}}{\text{ + 4OH}}^{\text{ - }}\to {\text{MnO}}_{\text{4}}^{\text{ - }}{\text{ + 3e}}^{\text{ - }}{\text{RHR:Ag}}^{\text{ + }}{\text{ + e}}^{\text{ - }}\to \text{Ag}$

 Balance atoms by adding ${\text{H}}_{\text{2}}\text{O}$to the opposite side.

${\text{OHR:MnO}}_{\text{2}}{\text{ + 4OH}}^{\text{ - }}\to {\text{MnO}}_{\text{4}}^{\text{ - }}{\text{ + 3e}}^{\text{ - }}{\text{ + 2H}}_{\text{2}}{\text{ORHR:Ag}}^{\text{ + }}{\text{ + e}}^{\text{ - }}\to \text{Ag}$

Multiply reactions by least common factor and combine half reactions, and cancel common ions/ compounds on both sides accordingly.

$\text{OHR }:{\text{MnO}}_2+4O{\text{H}}^{-}\to {\text{MnO}}_4^{-}+3e^{-}+2{\text{H}}_2\text{O}3xRHR:3{\text{Ag}}^{+}+3e^{-}\to 3\text{Ag Reaction }:2{\text{MnO}}_{2\left(s\right)}+4O{\text{H}}_{\left(aq\right)}^{-}+3{\text{Ag}}_{\left(aq\right)}^{+}\to 3{\text{Ag}}_{\left(s\right)}+{\text{MnO}}_{4\left(aq\right)}^{-}+2{\text{H}}_2{\text{O}}_{\left(l\right)}$

${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{MnO}}_{\text{4}}^{\text{ - }}}^{\text{o}}{\text{ = 0.59VE}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{Ag}}^{\text{ + }}}^{\text{o}}{\text{ = 0.80VE}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}^{\text{o}}{\text{ oxidation = E}}_{{\text{Ag}}^{\text{ + }}}^{\text{o}}{\text{ - E}}_{{\text{MnO}}_{\text{4}}^{\text{ - }}}^{\text{o}}\text{ = 0.80 - 0.59 = 0.21V}$

${\text{E}}_{\text{cell}}^{\text{o}}\text{ = 0.21}$, spontaneous reaction

 Therefore, the work done 

(a)${\text{PbO}}_{2\left(s\right)}+S{O}_{4\left(aq\right)}^{2-}+2C{u}_{\left(aq\right)}^{+}+4H_{\left(aq\right)}^{+}\to {\text{PbSO}}_{4\left(s\right)}+2{\text{H}}_2{\text{O}}_{\left(l\right)}+2{\text{Cu}}_{\left(aq\right)}^{2+};E_{cell}^o=1.55\text{V},$         spontaneous reaction

(b) ${\text{H}}_2{\text{O}}_{2\left(aq\right)}+N{i}_{\left(aq\right)}^{2+}\to 2H_{\left(aq\right)}^{+}+N{i}_{\left(s\right)}+H_2{O}_{2\left(aq\right)};E_{cell}^o=-0.93,$ non-spontaneous reaction