Question

21.Balance
the
following
redox
reactions
by
ion
–
electron
method
:


a)
Cr2O7
(aq)2–
+
SO3
2-(g)
→
Cr3+
(aq)
+
SO42–
(aq)
(in
acidic
solution​

Answer 1

Answer:

The balanced equation is

#"Cr"_2"O"_7^"2-" + "3SO"_3^"2-" + "8H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"4H"_2"O"#

Explanation:

Step 1: Separate the skeleton equation into two half-reactions.

#"Cr"_2"O"_7^"2-" → "Cr"^"3+"#

#"SO"_3^"2-" → "SO"_4^"2-"#

Step 2: Balance all atoms other than #"H"# and #"O"#.

#"Cr"_2"O"_7^"2-" → "2Cr"^"3+"#

#"SO"_3^"2-" → "SO"_4^"2-"#

Step 3: Balance #"O"#.

Add enough #"H"_2"O"# molecules to balance #"O"#.

#"Cr"_2"O"_7^"2-" → "2Cr"^"3+" + 7"H"_2"O"#

#"SO"_3^"2-" + "H"_2"O" → "SO"_4^"2-"#

Step 4: Balance #"H"#.

Add enough #"H"^"+"# ions to balance #"H"#.

#"Cr"_2"O"_7^"2-" + "14H"^"+" → "2Cr"^"3+" + 7"H"_2"O"#

#"SO"_3^"2-" + "H"_2"O" → "SO"_4^"2-" + "2H"^"+"#

Step 5: Balance charge.

Add electrons to the side that needs more negative charge.

#"Cr"_2"O"_7^"2-" + "14H"^"+" +6"e"^"-" → "2Cr"^"3+" + 7"H"_2"O"#

#"SO"_3^"2-" + "H"_2"O" → "SO"_4^"2-" + "2H"^"+" + 2"e"^"-"#

Step 6: Equalize electrons transferred.

Multiply each half-reaction by numbers to get the lowest common multiple of electrons transferred.

#1 ×["Cr"_2"O"_7^"2-" + "14H"^"+" +6"e"^"-" → "2Cr"^"3+" + 7"H"_2"O"]#

#3 × ["SO"_3^"2-" + "H"_2"O" → "SO"_4^"2-" + "2H"^"+" + 2"e"^"-"]#

Step 7: Add the two half-reactions.

#"Cr"_2"O"_7^"2-" + stackrelcolor(blue)(8)(color(red)(cancel(color(black)(14))))"H"^"+" + color(red)(cancel(color(black)(6"e"^"-"))) → "2Cr"^"3+" + stackrelcolor(blue)(4)(color(red)(cancel(color(black)(7))))"H"_2"O"#

#ul(3"SO"_3^"2-" + color(red)(cancel(color(black)(3"H"_2"O"))) → "3SO"_4^"2-" + color(red)(cancel(color(black)("6H"^"+"))) + color(red)(cancel(color(black)(6"e"^"-")))color(white)(mmmm))#

#"Cr"_2"O"_7^"2-" + "3SO"_3^"2-" + "8H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"4H"_2"O"#

Step 8: Check mass balance.

#ulbb("Atom"color(white)(m)"On the left"color(white)(m)"On the right")#

#color(white)(ml)"Cr"color(white)(mmmmll)2color(white)(mmmmmmm)2#

#color(white)(ml)"O"color(white)(mmmmll)16color(white)(mmmmmml)16#

#color(white)(ml)"S"color(white)(mmmmmll)3color(white)(mmmmmmll)3#

#color(white)(ml)"H"color(white)(mmmmlm)8color(white)(mmmmmmll)8#

Step 9. Check charge balance

#ulbb(color(white)(m)"On the left"color(white)(m))color(white)(mll)ulbb("On the right")#

#- 2 -6 +8 =0color(white)(mm)+6 - 6 = 0#

The balanced equation is

#"Cr"_2"O"_7^"2-" + "3SO"_3^"2-" + "8H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"4H"_2"O"#