Q44P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance each skeleton reaction, calculate and state whether the reaction is spontaneous:

(a) ${\mathbf{\text{Ag(s) + Cu}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq)Ag}}}^{\mathbf{\text{ + }}}\mathbf{\text{(aq) + Cu(s)}}$

(b) ${\mathbf{\text{Cd(s) + Cr}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{7}}}^{\mathbf{\text{2 - }}}{\mathbf{\text{(aq)Cd}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + Cr}}}^{\mathbf{\text{3 + }}}\mathbf{\text{(aq)}}$  

c)    ${\mathbf{\text{Ni}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + Pb(s)Ni(s) + Pb}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}}$

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: Given standard oxidation and reduction equations of redox reaction

(a) To balance each reaction, divide into half reactions and balance the electrons and atoms on both sides.

${\text{Ag}}_{\text{(s)}}{\text{ + Cu}}_{\text{(aq)}}^{\text{2 + }}\to {\text{Ag}}_{\text{(aq)}}^{\text{ + }}{\text{ + Cu}}_{\text{(s)}}{\text{2xOHR:2Ag}}_{\text{(s)}}\to {\text{2Ag}}_{\text{(aq)}}^{\text{ + }}{\text{ + 2e}}^{\text{ - }}{\text{RHR:Cu}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Cu}}_{\text{(s)}}{\text{ Reaction :Cu}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2Ag}}_{\text{(s)}}\to {\text{Cu}}_{\text{(s)}}{\text{ + Ag}}_{\text{(aq)}}^{\text{ + }}$

${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Ag}}^{\text{ + }}}^{\text{o}}{\text{ = 0.80VE}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{Cu}}^{\text{2 + }}}^{\text{o}}{\text{ = 0.15VE}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Cu}}^{\text{2 + }}}^{\text{o}}{\text{ - E}}_{{\text{Ag}}^{\text{ + }}}^{\text{o}}{\text{ = 0.15 - 0.80 = - 0.65VE}}_{\text{cell }}^{\text{o}}\text{ = - 0.65V}$ non-spontaneous reaction

(b) to balance each reaction, divide into half reactions and balance the redox through the half-reaction method.

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}^{\text{2 - }}{\text{ + Cd}}_{\text{(s)}}\to {\text{Cd}}_{\text{(aq)}}^{\text{2 + }}{\text{ + Cr}}_{\text{(aq)}}^{\text{3 + }}$

 Separate the half reactions.

 $\text{OHR:Cd}\to {\text{Cd}}^{\text{2 + }}{\text{RHR:Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}\to {\text{ + Cr}}^{\text{3 + }}$

 Balance electrons and non-O atoms.

 $\text{OHR:Cd}\to {\text{Cd}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}{\text{RHR:Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 6e}}^{\text{ - }}\to {\text{ + 2Cr}}^{\text{3 + }}$

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

 $\text{OHR:Cd}\to {\text{Cd}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}{\text{RHR:Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 6e}}^{\text{ - }}{\text{ + 14H}}^{\text{ + }}\to {\text{ + 2Cr}}^{\text{3 + }}$

 Balance atoms by adding $H_{2}O$ to the opposite side.

   $\text{OHR:Cd}\to {\text{Cd}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}{\text{RHR:Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 6e}}^{\text{ - }}{\text{ + 14H}}^{\text{ + }}\to {\text{ + 2Cr}}^{\text{3 + }}{\text{ + 7H}}_{\text{2}}\text{O}$$\text{OHR:Cd}\to {\text{Cd}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}{\text{RHR:Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 6e}}^{\text{ - }}{\text{ + 14H}}^{\text{ + }}\to {\text{ + 2Cr}}^{\text{3 + }}{\text{ + 7H}}_{\text{2}}\text{O}$

Step 2: Find a multiple reactions

 

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

$3xOHR:3\text{Cd}\to 3{\text{Cd}}^{2+}+2e^{-}RHR:{\text{Cr}}_2{\text{O}}_7^{2-}+6e^{-}+14{\text{H}}^{+}\to +2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O Reaction }:{\text{Cr}}_2{\text{O}}_{7\left(aq\right)}^{2-}+3{\text{Cd}}_{\left(s\right)}+14{\text{H}}_{\left(aq\right)}^{+}\to +2{\text{Cr}}_{\left(aq\right)}^{3+}+7{\text{H}}_2\text{O}\left(l\right)+3{\text{Cd}}_{\left(aq\right)}^{2+}$${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Cd}}^{\text{2 + }}}^{\text{o}}{\text{ = - 0.40VE}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{C}}_{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{\text{2 - }}}^{\text{o}}{\text{ = 1.33VE}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}^{\text{o}}{\text{ oxidation = E}}_{{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}}^{\text{o}}{\text{ - E}}_{{\text{Cd}}^{\text{2 + }}}^{\text{o}}{\text{ = 1.33 - ( - 0.40) = 1.74VE}}_{\text{cell }}^{\text{o}}\text{ = 1.74}$ To balance each reaction, divide into half reactions and balance the electrons and atoms on both sides.

 $N{i}_{\left(aq\right)}^{2+}+P{b}_{\left(s\right)}\to P{b}_{\left(aq\right)}^{2+}+N{i}_{\left(s\right)}$

 $RHR:N{i}_{\left(aq\right)}^{2+}+2e^{-}\to N{i}_{\left(s\right)}OHR:P{b}_{\left(s\right)}\to P{b}_{\left(aq\right)}^{2+}+2e^{-}$

 $\text{Reaction }:N{i}_{\left(aq\right)}^{2+}+P{b}_{\left(s\right)}\to P{b}_{\left(aq\right)}^{2+}+N{i}_{\left(s\right)}$

 ${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Pb}}^{\text{2 + }}}^{\text{o}}\text{ = - 0.13}$

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

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

 $=-0.25-(-0.13)=-0.12V$

 ${\text{E}}_{\text{cell }}^{\text{o}}\text{ = - 0.12,}$non-spontaneous reaction

 

Therefore ,the work done is

$\left(\text{a}\right){\text{Cu}}_{\left(aq\right)}^{2+}+2{\text{Ag}}_{\left(\text{s}\right)}\to \text{Cu}{u}_{\left(s\right)}+{\text{Ag}}_{\left(aq\right)}^{+};E_{cell}^o=-0.65\text{V},\text{ non - spontaneous reaction}$

$\left(b\right){\text{Cr}}_2{\text{O}}_{7\left(aq\right)}^{2-}+3{\text{Cd}}_{\left(s\right)}+14{\text{H}}_{\left(aq\right)}^{+}\to +2C{r}_{\left(aq\right)}^{3+}+7{\text{H}}_2\text{O}\left(l\right)+3C{d}_{\left(aq\right)}^{2+};E_{cell}^o=1.74\text{, spontaneous reaction}$ $\text{ (c) }N{i}_{\left(aq\right)}^{2+}+P{b}_{\left(s\right)}\to P{b}_{\left(aq\right)}^{2+}+N{i}_{\left(s\right)};E_{cell}^o=-0.12\text{, non - spontaneous reaction }$

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