# Question 86362

Jan 12, 2016

The balanced equation is

$\text{IO"_3^(-) + 6"Ag" + 6"H"^+ → "AgI"+ 5"Ag"^+ + 3"H"_2"O}$

#### Explanation:

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

We know that ${\text{IO}}_{3}^{-}$ is reduced to ${\text{I}}^{-}$ and $\text{Ag}$ is oxidized to ${\text{Ag}}^{+}$.

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

Step 1: Write the two half-reactions.

$\text{IO"_3^(-) → "AgI}$
${\text{Ag" → "Ag}}^{+}$

Step 2: Balance all atoms other than $\text{H}$ and $\text{O}$.

$\text{IO"_3^(-) + "Ag"^+ → "AgI}$
${\text{Ag" → "Ag}}^{+}$

Step 3: Balance $\text{O}$.

$\text{IO"_3^(-) + "Ag"^+ → "AgI" + color(red)(3)"H"_2"O}$
${\text{Ag" → "Ag}}^{+}$

Step 4: Balance $\text{H}$.

$\text{IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ → "AgI" + color(red)(3)"H"_2"O}$
${\text{Ag" → "Ag}}^{+}$

Step 5: Balance charge.

$\text{IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ + color(green)(6)"e"^(-)→ "AgI" + color(red)(3)"H"_2"O}$
${\text{Ag" → "Ag"^+ + color(magenta)(1)"e}}^{-}$

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.

$\text{IO"_3^(-) + color(red)(cancel(color(black)("Ag"^+))) + color(blue)(6)"H"^+ + color(red)(cancel(color(black)(color(green)(6)"e"^(-)))) → "AgI" + color(red)(3)"H"_2"O}$
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: $\text{1 I; 3 O; 6 Ag; 6 H}$
On the right: $\text{6 Ag; 1 I; 6 H; 3 O}$

Step 9: Check charge balance.

On the left: $\text{1- + 6+ = 5+}$
On the right: $\text{5+}$

∴ The balanced equation is

$\text{IO"_3^(-) + 6"Ag" + 6"H"^+ → "AgI"+ 5"Ag"^+ + 3"H"_2"O}$