Question

The number of moles of oxalate ions oxidized by one mole of permanganate ions in acidic medium is

• A. 2.5
• B. 5.0
• C. 1.5
• D. 2.0

Hint:

In redox reactions, the concept of equivalents is very powerful. The number of equivalents of the oxidizing agent must equal the number of equivalents of the reducing agent. Equivalents = Moles $\times$ n-factor. The n-factor is the number of electrons transferred per mole of the substance.

Answer 1

The Correct Option is A

This is a redox titration problem. We need to write the balanced chemical equation for the reaction between permanganate ions (MnO$_4^-$) and oxalate ions (C$_2$O$_4^{2-}$) in an acidic medium.
Step 1: Write the half-reactions.
Reduction half-reaction (permanganate):
In acidic medium, MnO$_4^-$ is reduced to Mn$^{2+}$. The oxidation state of Mn changes from +7 to +2.
The change in oxidation state is 5. So, the n-factor for MnO$_4^-$ is 5.
MnO$_4^-$ + 8H$^+$ + 5e$^-$ $\rightarrow$ Mn$^{2+}$ + 4H$_2$O.
Oxidation half-reaction (oxalate):
Oxalate ion (C$_2$O$_4^{2-}$) is oxidized to carbon dioxide (CO$_2$). The oxidation state of carbon changes from +3 (in C$_2$O$_4^{2-}$) to +4 (in CO$_2$).
For the two carbon atoms in one oxalate ion, the total change in oxidation state is $2 \times (+4 - +3) = 2$.
So, the n-factor for C$_2$O$_4^{2-}$ is 2.
C$_2$O$_4^{2-}$ $\rightarrow$ 2CO$_2$ + 2e$^-$.
Step 2: Balance the electrons and combine the half-reactions.
To balance the electrons, we multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.
$2 \times ($MnO$_4^-$ + 8H$^+$ + 5e$^-$ $\rightarrow$ Mn$^{2+}$ + 4H$_2$O)
$5 \times ($C$_2$O$_4^{2-}$ $\rightarrow$ 2CO$_2$ + 2e$^-$)
This gives:
2MnO$_4^-$ + 16H$^+$ + 10e$^-$ $\rightarrow$ 2Mn$^{2+}$ + 8H$_2$O
5C$_2$O$_4^{2-}$ $\rightarrow$ 10CO$_2$ + 10e$^-$
Adding them together gives the overall balanced equation:
2MnO$_4^-$ + 5C$_2$O$_4^{2-}$ + 16H$^+$ $\rightarrow$ 2Mn$^{2+}$ + 10CO$_2$ + 8H$_2$O.
Step 3: Determine the molar ratio.
From the balanced equation, we can see that 2 moles of permanganate ions react with 5 moles of oxalate ions.
The question asks for the number of moles of oxalate ions oxidized by one mole of permanganate ions.
Moles of oxalate = 1 mole MnO$_4^- \times \frac{5 \text{ moles } C_2O_4^{2-}}{2 \text{ moles } MnO_4^-} = \frac{5}{2} = 2.5$ moles.
Alternatively, using the concept of equivalents:
Equivalents of MnO$_4^-$ = Equivalents of C$_2$O$_4^{2-}$ Moles$_1 \times n$-factor$_1$ = Moles$_2 \times n$-factor$_2$ $1 \times 5 = \text{Moles}_{oxalate} \times 2$ Moles$_{oxalate} = 5/2 = 2.5$.