Question
Balance the following reactions by oxidation number method
✓
Answer
i. $Cr _2 O _{7( aq )}^{2-}+ SO _{3( aq )}^{2-} \longrightarrow Cr _{( aq )}^{3+}+ SO _{4( aq )}^{2-} \quad$ ( acidic )
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$
Cr _2 O _{7(a q)}^{2-}+ SO _{3(a)}^{2-} \longrightarrow 2 Cr _{( aq )}^{3+}+ SO _{4( aq )}^{2-}
$
Step 2: Assign oxidation number to $Cr$ and $S$. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 3 atoms of $S$ and 2 atoms of $Cr$. (There are already $2 Cr$ atoms.)
Step 3: Balance ' $O$ ' atoms by adding $4 H _2 O$ to the right-hand side.
$
Cr _2 O _{7( aq )}^{2-}+3 SO _{3( aq )}^{2-} \longrightarrow 2 Cr _{( aq )}^{3+}+3 SO _{4( aq )}^{2-}+4 H _2 O _{(l)}
$
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add $8 H$ on the left-hand side.
$
Cr _2 O _{7( aq )}^{2-}+3 SO _{3( aq )}^{2-}+8 H _{( aq )}^{+} \longrightarrow 2 Cr _{( aq )}^{3+}+3 SO _{4( aq )}^{2-}+4 H _2 O _{(l)}
$
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:
ii. $MnO _{4( aq )}^{-}+ Br _{( aq )}^{-} \longrightarrow MnO _{2( s )}+ BrO _3^{-}(a q) \quad$ (basic)
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$MnO _{4( aq )}^{-}+ Br _{( aq )}^{-} \longrightarrow MnO _{2( s )}+ BrO _{3( aq )}^{-}$
Step 2: Assign oxidation number to $Mn$ and Br. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 2 atoms of $Mn$.
$2 MnO _{4( aq )}^{-}+ Br _{( aq )}^{-} \longrightarrow 2 MnO _{2( s )}+ BrO _{3( aq )}^{-}$
Step 3: Balance ' $O ^{\text {at }}$ atoms by adding $H _2 O$ to the right-hand side.
$2 MnO _{4(a q)}^{-}+ Br _{(2 q)}^{-} \longrightarrow 2 MnO _{2( s )}+ BrO _{3( aq )}^{-}+ H _2 O _{(l)}$
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add $2 H ^{+}$on the left-hand side.
iii. $H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+ SO _{2( g )}+ H _2 O _{( l )}$ (acidic)
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$
H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+ SO _{2(g)}+ H _2 O _{( l )}
$
Step 2: Assign oxidation number to $S$ and $C$. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 2 atoms of $S$.
$
2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+2 SO _{2( g )}+ H _2 O _{( l )}
$
Step 3: Balance ' $O ^{\prime}$ atoms by adding $H 2 O$ to the right-hand side.
$
2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+2 SO _{2( g )}+ H _2 O _{( l )}+ H _2 O _{( l )}
$
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
$
2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _2+2 SO _{2( g )}+ H _2 O _{( l )}
$
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: $2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+2 SO _{2( g )}+ H _2 O _{( l )}$
iv. $Bi ( OH )_{3( s )}+ Sn ( OH )_{3( aq )}^{-} \longrightarrow Bi _{( s )}+ Sn ( OH )_{6( aq )}^{2-}$ (basic)
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$
Bi ( OH )_{3( s )}+ Sn ( OH )_{3( aq )}^{-} \longrightarrow Bi _{( s )}+ Sn ( OH )_{6( aq )}^{2-}
$
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.
Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.
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