Question

$n_1Cr_2O_7^{-2}+n_2FeC_2O_4\xrightarrow{Acidic\ Medium}n_3Cr^{+3}+n_4Fe^{+3}+n_5CO_2+n_6H_2O$

Consider the above redox reaction where $n_1, n_2, n_3, n_4, n_5$ and $n_6$ are the simplest non fractional coefficients of the balanced redox reaction. Then the value of $[n_1+n_2+n_3+n_4+n_5+n_6]$ is _____.

Answer 1

18

Answer 2

Solution:

1. Write the half‐reactions:

   • Reduction:

     $$\ce{Cr2O7^{2-} + 14 H^+ + 6 e^- -> 2 Cr^{3+} + 7 H2O}$$

   • Oxidation (for the oxalate ion):

     $$\ce{C2O4^{2-} -> 2 CO2 + 2 e^-}$$

   Also, note that in $\ce{FeC2O4}$ the Fe is in the +2 state and is oxidized to Fe$^{3+}$ (losing 1 electron). Therefore, per molecule of $\ce{FeC2O4}$:

   • Oxidation due to Fe: $\ce{Fe^{2+} -> Fe^{3+} + e^-}$
   • Oxidation due to oxalate: $\ce{C2O4^{2-} -> 2 CO2 + 2e^-}$

   Total electrons lost per $\ce{FeC2O4}$ = $1+2 = 3$ electrons.

2. Electron Balance:

   The reduction half-reaction requires 6 electrons. Therefore, we need 2 molecules of $\ce{FeC2O4}$ (since $2 \times 3 = 6$) to supply the 6 electrons.

3. Write the overall balanced reaction:

   $$\ce{Cr2O7^{2-} + 14 H^+ + 2 FeC2O4 -> 2 Cr^{3+} + 2 Fe^{3+} + 4 CO2 + 7 H2O}$$

4. Identify the coefficients:

   • $n_1 = 1$ (for $\ce{Cr2O7^{2-}}$)
   • $n_2 = 2$ (for $\ce{FeC2O4}$)
   • $n_3 = 2$ (for $\ce{Cr^{3+}}$)
   • $n_4 = 2$ (for $\ce{Fe^{3+}}$)
   • $n_5 = 4$ (for $\ce{CO2}$)
   • $n_6 = 7$ (for $\ce{H2O}$)

   Sum:

   $$n_1+n_2+n_3+n_4+n_5+n_6 = 1+2+2+2+4+7 = 18.$$

Short Explanation:

• Balance the reduction half-reaction for $\ce{Cr2O7^{2-}}$ (6e⁻ needed) and the oxidation half-reaction for the oxalate portion (2e⁻) plus Fe oxidation (1e⁻) in $\ce{FeC2O4}$ (total 3e⁻ per molecule).
• Use 2 molecules of $\ce{FeC2O4}$ to supply 6e⁻.
• Write the overall balanced equation and sum the coefficients.