Question #86362

1 Answer
Jan 12, 2016

Answer:

The balanced equation is

#"IO"_3^(-) + 6"Ag" + 6"H"^+ → "AgI"+ 5"Ag"^+ + 3"H"_2"O"#

Explanation:

You can find the general technique for balancing redox equations in acid solution here.

We know that #"IO"_3^-# is reduced to #"I"^-# and #"Ag"# is oxidized to #"Ag"^+#.

We also know that #"Ag"^+"(aq)" + "I"^(-)"(aq)" → "AgI(s)"#, so the iodide will exist as a precipitate of silver iodide.

Step 1: Write the two half-reactions.

#"IO"_3^(-) → "AgI"#
#"Ag" → "Ag"^+#

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

#"IO"_3^(-) + "Ag"^+ → "AgI"#
#"Ag" → "Ag"^+#

Step 3: Balance #"O"#.

#"IO"_3^(-) + "Ag"^+ → "AgI" + color(red)(3)"H"_2"O"#
#"Ag" → "Ag"^+#

Step 4: Balance #"H"#.

#"IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ → "AgI" + color(red)(3)"H"_2"O"#
#"Ag" → "Ag"^+#

Step 5: Balance charge.

#"IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ + color(green)(6)"e"^(-)→ "AgI" + color(red)(3)"H"_2"O"#
#"Ag" → "Ag"^+ + color(magenta)(1)"e"^(-)#

Step 6: Equalize electrons transferred.

#color(magenta)(1) × ["IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ + color(green)(6)"e"^(-)→ "AgI" + color(red)(3)"H"_2"O"]#
#color(green)(6) × ["Ag" → "Ag"^+ + color(magenta)(1)"e"^(-)]#

Step 7: Add the two half-reactions.

#"IO"_3^(-) + color(red)(cancel(color(black)("Ag"^+))) + color(blue)(6)"H"^+ + color(red)(cancel(color(black)(color(green)(6)"e"^(-)))) → "AgI" + color(red)(3)"H"_2"O"#
#6"Ag" → stackrel(5)(color(red)(cancel(color(black)(6))))"Ag"^+ + color(red)(cancel(color(black)(color(magenta)(6)"e"^(-))))#
#stackrel(———————————————————)("IO"_3^(-) + 6"Ag" + color(blue)(6)"H"^+ → "AgI"+ 5"Ag"^+ + color(red)(3)"H"_2"O")#

Step 8: Check mass balance.

On the left: #"1 I; 3 O; 6 Ag; 6 H"#
On the right: #"6 Ag; 1 I; 6 H; 3 O"#

Step 9: Check charge balance.

On the left: #"1- + 6+ = 5+"#
On the right: #"5+"#

∴ The balanced equation is

#"IO"_3^(-) + 6"Ag" + 6"H"^+ → "AgI"+ 5"Ag"^+ + 3"H"_2"O"#