How many electrons are involved in the following redox reaction?
Cr$$_2$$O$$_7^{2-}$$ + Fe$$^{2+}$$ + C$$_2$$O$$_4^{2-}$$ → Cr$$^{3+}$$ + Fe$$^{3+}$$ + CO$$_2$$ (Unbalanced)

To determine the number of electrons involved in the redox reaction, we first need to balance the reaction since it is given unbalanced. The reaction is:

$$\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} + \text{CO}_2$$

We start by identifying the oxidation states of each element:

• In $$\text{Cr}_2\text{O}_7^{2-}$$, oxygen is $$-2$$. Let the oxidation state of Cr be $$x$$. Then $$2x + 7(-2) = -2$$, so $$2x - 14 = -2$$, giving $$2x = 12$$, and $$x = +6$$. Thus, Cr is $$+6$$.
• In $$\text{Fe}^{2+}$$, iron is $$+2$$.
• In $$\text{C}_2\text{O}_4^{2-}$$, oxygen is $$-2$$. Let the oxidation state of C be $$y$$. Then $$2y + 4(-2) = -2$$, so $$2y - 8 = -2$$, giving $$2y = 6$$, and $$y = +3$$. Thus, each carbon is $$+3$$.
• In $$\text{Cr}^{3+}$$, chromium is $$+3$$.
• In $$\text{Fe}^{3+}$$, iron is $$+3$$.
• In $$\text{CO}_2$$, oxygen is $$-2$$, so carbon is $$+4$$.

Now, observe the changes in oxidation states:

• Chromium is reduced from $$+6$$ to $$+3$$, gaining electrons.
• Iron is oxidized from $$+2$$ to $$+3$$, losing electrons.
• Carbon in oxalate is oxidized from $$+3$$ to $$+4$$, losing electrons.

We split the reaction into half-reactions:

Reduction: $$\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$$

Oxidation: $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$$ and $$\text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2$$

Balance the reduction half-reaction in acidic medium:

1. Balance Cr atoms: $$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$$.
2. Balance oxygen by adding water: $$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$.
3. Balance hydrogen by adding H⁺: $$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$.
4. Balance charge: Left side charge is $$-2 + 14(+1) = +12$$, right side is $$2(+3) = +6$$. Add 6 electrons to the left: $$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$.

So, reduction involves 6 electrons per $$\text{Cr}_2\text{O}_7^{2-}$$ ion.

Balance the oxidation half-reactions:

• For iron: $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$. Each $$\text{Fe}^{2+}$$ loses 1 electron.
• For oxalate: $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$$. Oxidation state change: Two carbons go from $$+3$$ to $$+4$$, so total loss of 2 electrons. Balance atoms and charge: Left has charge $$-2$$, right has charge $$0$$. Add 2 electrons to the right: $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$$. Each $$\text{C}_2\text{O}_4^{2-}$$ loses 2 electrons.

Now, combine the half-reactions. Let the number of $$\text{Cr}_2\text{O}_7^{2-}$$ ions be $$a$$, $$\text{Fe}^{2+}$$ ions be $$b$$, and $$\text{C}_2\text{O}_4^{2-}$$ ions be $$c$$.

Electron balance: Electrons gained = electrons lost. Reduction gains $$6a$$ electrons. Oxidation loses $$b$$ electrons from iron and $$2c$$ electrons from oxalate. So:

$$6a = b + 2c$$

The balanced reaction must include H⁺ and H₂O. Using the reduction half-reaction, for each $$\text{Cr}_2\text{O}_7^{2-}$$, we need 14H⁺ and produce 7H₂O. The full reaction is:

$$a \text{Cr}_2\text{O}_7^{2-} + b \text{Fe}^{2+} + c \text{C}_2\text{O}_4^{2-} + 14a \text{H}^+ \rightarrow 2a \text{Cr}^{3+} + b \text{Fe}^{3+} + 2c \text{CO}_2 + 7a \text{H}_2\text{O}$$

Balance atoms and charge:

• Oxygen: Left has $$7a + 4c$$ atoms (from $$\text{Cr}_2\text{O}_7^{2-}$$ and $$\text{C}_2\text{O}_4^{2-}$$), right has $$7a$$ from H₂O and $$4c$$ from CO₂ (since each CO₂ has 2 oxygen atoms, and $$2c$$ CO₂ has $$4c$$ oxygen atoms). So $$7a + 4c = 7a + 4c$$, balanced.
• Hydrogen: Left has $$14a$$ H atoms from H⁺, right has $$14a$$ H atoms from $$7a$$ H₂O, balanced.
• Charge: Left charge is $$-2a + 2b - 2c + 14a = 12a + 2b - 2c$$. Right charge is $$2a \times 3 + b \times 3 = 6a + 3b$$. Set equal: $$12a + 2b - 2c = 6a + 3b$$, simplifying to $$6a - b - 2c = 0$$, or $$b + 2c = 6a$$, which matches the electron balance equation.

Solve $$b + 2c = 6a$$ for positive integers $$a$$, $$b$$, $$c$$. Since the unbalanced reaction shows one of each reactant, we find minimal solutions:

• If $$a = 1$$, then $$b + 2c = 6$$. Possible pairs: $$(b, c) = (4, 1)$$ or $$(2, 2)$$.

Both are valid balanced equations:

1. For $$(a, b, c) = (1, 4, 1)$$: $$\text{Cr}_2\text{O}_7^{2-} + 4\text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 4\text{Fe}^{3+} + 2\text{CO}_2 + 7\text{H}_2\text{O}$$
2. For $$(a, b, c) = (1, 2, 2)$$: $$\text{Cr}_2\text{O}_7^{2-} + 2\text{Fe}^{2+} + 2\text{C}_2\text{O}_4^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 2\text{Fe}^{3+} + 4\text{CO}_2 + 7\text{H}_2\text{O}$$

In both cases, the total electrons transferred are 6, as reduction gains 6 electrons and oxidation loses 6 electrons:

• Case 1: Iron loses $$4 \times 1 = 4$$ electrons, oxalate loses $$1 \times 2 = 2$$ electrons, total 6.
• Case 2: Iron loses $$2 \times 1 = 2$$ electrons, oxalate loses $$2 \times 2 = 4$$ electrons, total 6.

The number of electrons involved is always 6 per mole of $$\text{Cr}_2\text{O}_7^{2-}$$, regardless of the coefficients chosen for the reducing agents.

Hence, the correct answer is Option A.