Question

Consider the following reaction, $xMnO_4^- + yC_2O_4^{2-}+zH^+ \rightarrow xMn^{2+}+2yCO_2 + \frac{z}{2}H_2O$ The values of x, y and z in the reaction are, respectively

• A. 5.2 and 16
• B. 2,5 and 8
• C. 2,5 and 16
• D. 5.2 and 8

Answer 1

The Correct Option is C

 To determine the values of \(x\), \(y\), and \(z\) in the given redox reaction, we first need to ensure that the reaction is balanced both in terms of mass and charge.

The given reaction is:

\(xMnO_4^- + yC_2O_4^{2-}+zH^+ \rightarrow xMn^{2+}+2yCO_2 + \frac{z}{2}H_2O\)

1. Begin by identifying the oxidation states and the changes involved:
   • \(MnO_4^-\rightarrow Mn^{2+}\): The manganese is reduced from +7 to +2, a reduction of 5 electrons per manganese.
   • \(C_2O_4^{2-} \rightarrow CO_2\): Each carbon atom is oxidized from +3 in \(C_2O_4^{2-}\) to +4 in CO2, losing 2 electrons per oxalate ion.
2. Balancing the redox reaction:
   • To balance the electrons, note that 2 moles of \(MnO_4^-\\) will accept 10 electrons (2 * 5 = 10).
   • Thus, 5 moles of \(\text{C}_2O_4^{2-}\) will donate those 10 electrons (5 * 2 = 10).
3. Thus, \(x = 2\) and \(y = 5\).
4. Next, balance the charges using hydrogen ions \((H^+)\):
   • The left side: 2 \(MnO_4^-\\) ions contribute a charge of 2 * (-1) = -2.
   • The left side: 5 \(C_2O_4^{2-}\) ions contribute a charge of 5 * (-2) = -10.
   • Adding them, we get a total negative charge of -12 on the left.
   • On the right-hand side, we have 2 \(Mn^{2+}\) ions contributing +4.
   • Hence, we need \(z\) protons \((H^+)\) to balance the charge.
   • Therefore, -12 + \(z\) = +4, giving us \(z = 16\).
5. The final balanced reaction is:
6. Therefore, the values of \(x\), \(y\), and \(z\) are 2, 5, and 16, respectively.

The correct option is: 2, 5, and 16.