Question

a) \\(\text{cr}_2\text{o}_7^{-2} + \text{cl}_2 \
ightarrow \text{clo}_3^{-1} + \text{cr}^{+3}\\)

b) \\(\text{fe}(\text{oh})_2 + \text{cro}_4^{-2} \
ightarrow \text{fe}(\text{oh})_3 + \text{cr}(\text{oh})_3\\) (basic)

Explanation:

Identify half-reactions for equation (a)

Using the Oxidation Numbers and Redox Rules knowledge points, we determine the oxidation states and separate the reaction into reduction and oxidation half-reactions:

• Reduction: \(\text{Cr}_2\text{O}_7^{2-}

ightarrow 2\text{Cr}^{3+}\)

• Oxidation: \(\text{Cl}_2

ightarrow 2\text{ClO}_3^-\)

Balance atoms and charges for (a)

Balance each half-reaction in acidic solution:

• Reduction:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- ightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

• Oxidation:

$$\text{Cl}_2 + 6\text{H}_2\text{O} ightarrow 2\text{ClO}_3^- + 12\text{H}^+ + 10e^-$$

Combine and simplify equation (a)

Multiply the reduction half-reaction by 5 and the oxidation half-reaction by 3 to equalize electrons (\(30e^-\)):

• \(5\text{Cr}_2\text{O}_7^{2-} + 70\text{H}^+ + 30e^-

ightarrow 10\text{Cr}^{3+} + 35\text{H}_2\text{O}\)

• \(3\text{Cl}_2 + 18\text{H}_2\text{O}

ightarrow 6\text{ClO}_3^- + 36\text{H}^+ + 30e^-\)

Combine and cancel common terms:

$$5\text{Cr}_2\text{O}_7^{2-} + 3\text{Cl}_2 + 34\text{H}^+ ightarrow 10\text{Cr}^{3+} + 6\text{ClO}_3^- + 17\text{H}_2\text{O}$$

Identify half-reactions for equation (b)

For the basic solution reaction:

• Oxidation: \(\text{Fe(OH)}_2

ightarrow \text{Fe(OH)}_3\)

• Reduction: \(\text{CrO}_4^{2-}

ightarrow \text{Cr(OH)}_3\)

Balance (b) in basic solution

Balance each half-reaction using \(\text{OH}^-\) and \(\text{H}_2\text{O}\):

• Oxidation:

$$\text{Fe(OH)}_2 + \text{OH}^- ightarrow \text{Fe(OH)}_3 + e^-$$

• Reduction:

$$\text{CrO}_4^{2-} + 4\text{H}_2\text{O} + 3e^- ightarrow \text{Cr(OH)}_3 + 5\text{OH}^-$$

Multiply the oxidation half-reaction by 3 and combine:

$$3\text{Fe(OH)}_2 + \text{CrO}_4^{2-} + 4\text{H}_2\text{O} ightarrow 3\text{Fe(OH)}_3 + \text{Cr(OH)}_3 + 2\text{OH}^-$$

Answer:

Question a

$$5\text{Cr}_2\text{O}_7^{2-} + 3\text{Cl}_2 + 34\text{H}^+ ightarrow 10\text{Cr}^{3+} + 6\text{ClO}_3^- + 17\text{H}_2\text{O}$$

Question b

$$3\text{Fe(OH)}_2 + \text{CrO}_4^{2-} + 4\text{H}_2\text{O} ightarrow 3\text{Fe(OH)}_3 + \text{Cr(OH)}_3 + 2\text{OH}^-$$