Answer:
(a) Reaction:
\[MnO_{4}^{-}(aq)+{{I}^{-}}(aq)\to
Mn{{O}_{2}}(s)+{{I}_{2}}(s)\]
1st step: Splitting
into two half reactions: (balance the elements other than hydrogen and oxygen)
\[MnO_{4}^{-}\to Mn{{O}_{2}}(s)\] (Reduction)
\[2{{I}^{-}}\to {{I}_{2}}(s)\] (Oxidation)
2nd step: Balance
oxygen atoms by adding water molecule to the oxygen deficient side.
\[MnO_{4}^{-}\to Mn{{O}_{2}}+2{{H}_{2}}O\]
\[2{{I}^{-}}\to
{{I}_{2}}\]
3rd step: Balance
hydrogen by adding \[{{H}^{+}}\] ions to the hydrogen deficient side.
\[MnO_{4}^{-}+4{{H}^{+}}\to
Mn{{O}_{2}}+2{{H}_{2}}O\]
\[2{{I}^{-}}\to {{I}_{2}}\]
4th step: Balance the
charge by adding electrons
\[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to
Mn{{O}_{2}}+2{{H}_{2}}O\] \[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}\]
5th step: Equations
are added in such a way that electrons are cancelled; we now get the balanced
equation.
\[[MnO_{4}^{-}]+4{{H}^{+}}+3{{e}^{-}}\to
Mn{{O}_{2}}+2{{H}_{2}}O]\times 2\]
\[[2{{I}^{-}}\to
{{I}_{2}}+2{{e}^{-}}]\times 3\]
\[\frac{\frac{{}}{2MnO_{4}^{-}+6{{I}^{-}}+8{{H}^{+}}\to
2Mn{{O}_{2}}+3{{I}_{2}}+4{{H}_{2}}O}}{Balanced\text{ }equation}\]
\[2MnO_{4}^{-}+4{{H}_{2}}O\to
2Mn{{O}_{2}}+3\underset{(Basic\,medium)}{\mathop{{{I}_{2}}+8O{{H}^{-}}}}\,\]
(b) The given equation is:
\[MnO_{4}^{-}(aq)+S{{O}_{2}}(g)\to
M{{n}^{2+}}(aq)+\underset{\left( acidic\text{ }medium
\right)}{\mathop{HSO_{4}^{-}(aq)}}\,\] Step I: Splitting into two half
equations:
\[MnO_{4}^{-}(aq)\to M{{n}^{2+}}(aq);\]
\[S{{O}_{2}}(g)\to HSO_{4}^{-}(aq)\]
Step II: Balancing oxygen by adding water molecules:
\[MnO_{4}^{-}(aq)\to
\overset{2+}{\mathop{Mn}}\,(aq)+4{{H}_{2}}O\]
\[S{{O}_{2}}(g)+2{{H}_{2}}O\to HSO_{4}^{-}(aq)\]
Step III: Balancing hydrogen by adding \[{{H}^{+}}\]ions:
\[MnO_{4}^{-}(aq)+8{{H}^{+}}\to
M{{n}^{2+}}(aq)+4{{H}_{2}}O\]
\[S{{O}_{2}}(g)+2{{H}_{2}}O\to HSO_{4}^{-}(aq)+3{{H}^{+}}\]
Step IV: Balancing the charge by adding electrons:
\[[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to
M{{n}^{2+}}+4{{H}_{2}}O]\times 2\]
\[[S{{O}_{2}}(g)+2{{H}_{2}}O\to
HSO_{4}^{-}(aq)+3{{H}^{+}}+2{{e}^{-}}]\times 5\]Step V: Adding above equations, we get:
\[2MnO_{4}^{-}(aq)+5S{{O}_{2}}(g)+2{{H}_{2}}O+{{H}^{+}}\]
\[\to
2M{{n}^{2+}}(aq)+5HSO_{4}^{-}(aq)\]
It is now balanced equation.
(c)\[{{H}_{2}}{{O}_{2}}(aq)+F{{e}^{2+}}(aq)\to
F{{e}^{3+}}(aq)+{{H}_{2}}O(l)\]
Step I: Splitting of equation
into two half equations:
\[F{{e}^{2+}}\to F{{e}^{3+}}\]
\[{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O\]
Step II: Oxygen and
other elements are balanced hydrogen will be balanced by adding \[{{H}^{+}}\]
ions to the hydrogen deficient side.
\[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}\]
\[{{H}_{2}}{{O}_{2}}+2{{H}^{+}}\to
2{{H}_{2}}O\]
Step III: Charge is
now balanced by adding electrons to the side where positive charge is excess or
negative charge is less.
\[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}\]
\[{{H}_{2}}{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\to 2{{H}_{2}}O\]
Step IV: Add these
equations in such a way that electrons are cancelled. The resultant equation
will now be a balanced equation.
\[\begin{align}
& 2[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}] \\
& \frac{{{H}_{2}}{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\to
2{{H}_{2}}O}{\frac{2F{{e}^{2+}}+{{H}_{2}}{{O}_{2}}+2{{H}^{+}}\to
2F{{e}^{3+}}+2{{H}_{2}}O}{{}}} \\
\end{align}\]
(d) \[C{{r}_{2}}O_{7}^{2-}+S{{O}_{2}}+{{H}^{+}}\to
C{{r}^{3+}}+HSO_{4}^{-}+{{H}_{2}}O\]
Step I: Splitting into
two half reactions.
\[\underset{(Reduction\,half\,reaction)}{\mathop{C{{r}_{2}}O_{7}^{2-}+{{H}^{+}}\to
C{{r}^{3+}}+{{H}_{2}}O}}\,\]\[\underset{(OXidation\,half\,reaction)}{\mathop{S{{O}_{2}}\to
HSO_{4}^{-}}}\,\]
Step II: Adding \[{{H}^{+}}\]
ions to the side deficient in hydrogen.
\[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\to
C{{r}^{3+}}+7{{H}_{2}}O\]
Step III: Adding water
to the side deficient in oxygen,
\[S{{O}_{2}}+2{{H}_{2}}O\to
HSO_{4}^{-}+3{{H}^{+}}\]
Step IV: Making atoms
equal on both sides,
\[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\to
2C{{r}^{3+}}+7{{H}_{2}}O\]
Step V: Adding
electrons to the sides deficient in electrons,
\[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6e\to
2C{{r}^{3+}}+7{{H}_{2}}O;\] \[S{{O}_{2}}+2{{H}_{2}}O\to
HSO_{4}^{-}+3{{H}^{+}}+2e\]
Step VI: Balancing
electrons in both the half reaction,
\[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6e\to
2C{{r}^{3+}}+7{{H}_{2}}O\]
\[[S{{O}_{2}}+2{{H}_{2}}O\to
HSO_{4}^{-}+3{{H}^{+}}+2e]\times 3\]Step VII: Adding both the half reactions,
\[C{{r}_{2}}O_{7}^{2-}+5{{H}^{+}}+3S{{O}_{2}}\to
2C{{r}^{3+}}+3HSO_{4}^{-}+{{H}_{2}}O\]
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