Question

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

(a)${\mathbf{\text{Co(s) +H}}}^{\mathbf{\text{+}}}{\mathbf{\text{(aq)Co}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq) +H}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$ (b) ${\mathbf{\text{Mn}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq) + Br}}}_{\mathbf{\text{2}}}{\mathbf{\text{(l)nMnO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{-}}}{\mathbf{\text{(aq) + Br}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}$

(c) $\mathbf{\text{H}}{{\mathbf{\text{g}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq)Hg}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq) + Hg(l)}}$

Short answer

The required work is used to acidic solutions react spontaneously process of reduction and oxidation also used into half reaction and the electrons and atoms on both sides.

Step-by-step solution

 Electrons and atoms

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

${\text{OHR :Co}}_{\text{(s)}}\to {\text{Co}}_{\text{(aq)}}^{\text{2 +}}{\text{+ 2e}}^{\text{-}}$

${\text{RHR:2H}}_{\text{(aq)}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2(g)}}$

${\text{Reaction:2H}}_{\text{(aq)}}^{\text{+}}{\text{+ Co}}_{\text{(s)}}\to {\text{H}}_{\text{2(g)}}{\text{+ Co}}_{\text{(aq)}}^{\text{2 +}}$

Given standard reduction equations

$$\begin{gathered} {\text{E}}_{\text{oxidation}}^{\text{o}}{\text{= E}}_{{\text{C0}}^{\text{2 +}}}^{\text{o}}\text{=-0.28V} \\ {\text{E}}_{\text{reduction}}^{\text{o}}{\text{= E}}_{{\text{H}}^{\text{+}}}^{\text{o}}\text{=0V} \\ {\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{reduction}}^{\text{o}}{\text{- E}}_{\text{oxidation}}^{\text{o}} \\ {\text{= E}}_{{\text{H}}^{\text{+}}}^{\text{o}}{\text{- E}}_{{\text{Co}}^{\text{2 +}}}^{\text{o}} \\ =0-(-0.28)=0.28V \end{gathered}$$

Finding redox half-reactions

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

${\text{Mn}}_{\text{(aq)}}^{\text{2 +}}{\text{+ Br}}_{\text{2(l)}}\to {\text{M}}_{\text{4(aq)}}^{\text{-}}{\text{+ Br}}_{\text{(aq)}}^{\text{-}}\text{[acidic ]}$

Separate the half reactions.

${\text{OHR:Mn}}^{\text{2 +}}\to {\text{MnO}}_{\text{4}}^{\text{-}}{\text{RHR:Br}}_{\text{2}}\to {\text{Br}}^{\text{-}}$

Balance electrons and non-O atoms.

${\text{OHR:Mn}}^{\text{2 +}}\to {\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5e}}^{\text{-}}{\text{RHR:Br}}_{\text{2}}{\text{+ 2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}$

Balance reaction charges with

${\text{H}}^{\text{+}}{\text{.OHR:Mn}}^{\text{2 +}}\to {\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5e}}^{\text{-}}{\text{+ 8H}}^{\text{+}}{\text{RHR:Br}}_{\text{2}}{\text{+ 2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}$

Balance${\text{H}}_{\text{2}}\text{O}$ to the opposite side

${\text{OHR:4H}}_{\text{2}}{\text{O + Mn}}^{\text{2 +}}\to {\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5e}}^{\text{-}}{\text{+ 8H}}^{\text{+}}{\text{RHR:Br}}_{\text{2}}{\text{+ 2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}$

Given multiple reactions

Multiply reactions by least common factor and combine half reactions.

${\text{2xOHR:8H}}_{\text{2}}{\text{O + 2Mn}}^{\text{2 +}}\to {\text{2MnO}}_{\text{4}}^{\text{-}}{\text{+ 10e}}^{\text{-}}{\text{+ 16H}}^{\text{+}}{\text{5xRHR:5Br}}_{\text{2}}{\text{+ 10e}}^{\text{-}}\to {\text{10Br}}^{\text{-}}$

${\text{8H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+ 2Mn}}_{\text{(aq)}}^{\text{2 +}}{\text{+ 5Br}}_{\text{2(l)}}\to {\text{2MnO}}_{\text{4(aq)}}^{\text{-}}{\text{+ 10B}}_{\text{(aq)}}^{\text{-}}{\text{+ 16H}}_{\text{(aq)}}^{\text{+}}$

 Find spontaneous and non-spontaneous reaction

${\text{E}}_{\text{oxidation}}^{\text{o}}{\text{= E}}_{{\text{MnO}}_{\text{4}}^{\text{-}}}^{\text{o}}\text{= 1.51}$

${\text{E}}_{\text{reduction}}^{\text{o}}{\text{= E}}_{{\text{Br}}_{\text{2}}}^{\text{o}}\text{= 1.07}$

$\begin{array}{rcl}& & {\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{reduction}}^{\text{o}}{\text{- E}}_{\text{oxidation}}^{\text{o}}\\ & & {\text{= E}}_{{\text{Br}}_{\text{2}}}^{\text{o}}{\text{- E}}_{{\text{MnO}}_{\text{4}}^{\text{-}}}^{\text{o}}\\ & & \text{=1.07-1.51=-0.44V}\\ & & \end{array}$

${\text{E}}_{\text{cell}}^{\text{o}}\text{= - 0.44}$, non-spontaneous reaction

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

${\text{Hg}}_{\text{2(aq)}}^{\text{2 +}}\to {\text{Hg}}_{\text{(aq)}}^{\text{2 +}}{\text{+ Hg}}_{\text{(l)}}$
${\text{OHR:Hg}}_{\text{2(aq)}}^{\text{2 +}}\to {\text{2Hg}}_{\text{(aq)}}^{\text{2 +}}{\text{+ 2e}}^{\text{-}}$

${\text{RHR:Hg}}_{\text{2(aq)}}^{\text{2 +}}{\text{+ 2e}}^{\text{-}}\to {\text{2Hg}}_{\text{(l)}}$

${\text{Reaction :2Hg}}_{\text{2(aq)}}^{\text{2 +}}\to {\text{2Hg}}_{\text{(aq)}}^{\text{2 +}}{\text{+ 2Hg}}_{\text{(l)}}$

${\text{E}}_{\text{oxidation}}^{\text{o}}{\text{= E}}_{{\text{Hg}}^{\text{2 +}}}^{\text{o}}{\text{= 0.92E}}_{\text{reduction}}^{\text{o}}{\text{= E}}_{{\text{Hg}}_{\text{2}}^{\text{2 +}}}^{\text{o}}\text{= 0.85}$

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

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

$\text{= 0.85 - 0.92}$

$=-0.07V$

Therefore, the work done is

(a)${\text{2H}}_{\text{(aq)}}^{\text{+}}{\text{+ Co}}_{\text{(s)}}\to {\text{H}}_{\text{2}}{\text{(g) + Co}}_{\text{(aq)}}^{\text{2 +}}{\text{;E}}_{\text{cell}}^{\text{o}}\text{=0.28V}$,spontaneous reaction

(b),${\text{8H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+ 2Mn}}_{\text{(aq)}}^{\text{2 +}}{\text{+ 5Br}}_{\text{2}}\text{(l)}\to {\text{2MnO}}_{\text{4(aq)}}^{\text{-}}{\text{+ 10Br}}_{\text{(aq)}}^{\text{-}}{\text{+ 16H}}_{\text{(aq)}}^{\text{+}}{\text{;E}}_{\text{cell}}^{\text{o}}\text{=-0.44}$

,non-spontaneous reaction

(c)${\text{2Hg}}_{\text{2(aq)}}^{\text{2 +}}\to {\text{2Hg}}_{\text{(aq)}}^{\text{2 +}}{\text{+ 2Hg}}_{\text{(l)}}{\text{;E}}_{\text{cell}}^{\text{o}}\text{=-0.07}$,non-spontaneous reaction