Consider the following reaction: xMnO4¯ + yC2O42– + zH+ → xMn2+ + 2yCO2 + z2 H2O The value of x, y and z in the reaction are, respectively:

Consider the following reaction: xMnO4¯ + yC2O42– + zH+ \overset{}{\rightarrow} xMn2+ + 2yCO2 + \frac{z}{2}H2O The value of x, y and z in the reaction are, respectively:

Engineering

Chemistry

Chemical Equation and Stoichiometric Calculation

Balancing of Redox Reaction

Empirical and Molecular Formula

Question

Consider the following reaction:

xMnO₄¯ + yC₂O₄^2– + zH⁺ $\overset{}{\to}$ xMn²⁺ + 2yCO₂ + $\frac{z}{2}$ H₂O

The value of x, y and z in the reaction are, respectively:

2, 5 and 16

5, 2 and 8

5, 2 and 16

2, 5 and 8

2MnO₄¯ + 5C₂O₄^2– + 16H⁺ $\overset{}{\to}$ 2Mn²⁺ + 10CO₂ + 8H₂O

This is a redox reaction balancing problem involving permanganate (MnO₄⁻) and oxalate (C₂O₄²⁻) ions in acidic medium. The key concept is that the number of electrons lost in oxidation must equal the number of electrons gained in reduction.

Step 1: Identify oxidation states and changes

For MnO₄⁻ → Mn²⁺:

Let oxidation state of Mn in MnO₄⁻ be $a$ . Then:

$a + 4 \times (- 2) = - 1$ ⇒ $a - 8 = - 1$ ⇒ $a = + 7$

In Mn²⁺, oxidation state is +2. So change is +7 → +2: gain of 5 electrons per Mn atom.

For C₂O₄²⁻ → 2CO₂:

Let oxidation state of C in C₂O₄²⁻ be $b$ . Then:

$2 b + 4 \times (- 2) = - 2$ ⇒ $2 b - 8 = - 2$ ⇒ $2 b = 6$ ⇒ $b = + 3$

In CO₂, oxidation state of C is +4. So each carbon changes from +3 to +4: loss of 1 electron per C atom.

Since one C₂O₄²⁻ has 2 carbon atoms, total electrons lost per oxalate ion = 2 electrons.

Step 2: Balance electron transfer

Let the number of MnO₄⁻ ions be $x$ and C₂O₄²⁻ ions be $y$ .

Total electrons gained by MnO₄⁻ = $5 x$

Total electrons lost by C₂O₄²⁻ = $2 y$

For balance: $5 x = 2 y$ ⇒ $y = \frac{5 x}{2}$

To avoid fractions, take $x = 2$ , then $y = 5$ .

Step 3: Balance atoms and charge

With $x = 2$ , $y = 5$ , the unbalanced reaction is:

$2 {MnO}_{4} ⁻ + 5 C_{2} O_{4}^{2 ⁻} + z H^{+} \to 2 {Mn}^{2 +} + 10 CO 2 + \frac{z}{2} H_{2} O$

Balance oxygen atoms: Left side: from MnO₄⁻ = 2×4 = 8, from C₂O₄²⁻ = 5×4 = 20, total = 28 oxygen atoms.

Right side: from CO₂ = 10×2 = 20, from H₂O = (z/2)×1. So: 20 + z/2 = 28 ⇒ z/2 = 8 ⇒ z = 16.

Verify charge balance: Left side: from MnO₄⁻ = 2×(-1) = -2, from C₂O₄²⁻ = 5×(-2) = -10, from H⁺ = 16×(+1) = +16. Total charge = -2 -10 +16 = +4.

Right side: from Mn²⁺ = 2×(+2) = +4. Total charge = +4. Balanced.

Final Answer

Thus, x = 2, y = 5, z = 16.

Detailed Solution

Step 1: Identify oxidation states and changes

Step 2: Balance electron transfer

Step 3: Balance atoms and charge

Final Answer

Related Topics & Formulae

Detailed Solution

Step 1: Identify oxidation states and changes

Step 2: Balance electron transfer

Step 3: Balance atoms and charge

Final Answer

Related Topics & Formulae

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