# Why does balancing the oxidation of dichromate(VI) to chromium(III) require six iron atoms?

The problem:

Balance the following chemical equation using oxidation numbers:

$$\ce{Fe^2+ + Cr_2O_7^2- -> Fe^3+ + Cr^3+}$$

What I have tried:

$$\ce{\overset{+II}{Fe}^{2+}(aq) + \overset{+VI}{Cr}_2O7^2–(aq) \rightarrow \overset{+III}{Fe}^3+(aq) + \overset{+III}{Cr}^3+(aq)}$$

\begin{align} \text{Electron transition:}\quad \ce{\overset{+VI}{Cr}_2 + 6 e- &-> 2 \overset{+III}{Cr^{3+}}} &\quad &|\times 1 \\ \ce{\overset{+II}{Fe} &-> \overset{+III}{Fe} + 1 e-} &\quad &|\times 3 \end{align}

$$\ce{3Fe^{2+}(aq) + Cr2O7^2–(aq) –> 2 Cr^3+(aq) + 3Fe^{3+}(aq)}$$ \begin{align} \text{Charges:}\quad &\text{left side} &\quad 3\cdot(+2) + 1\cdot(-2) &= +4 &\quad &|\text{add}~\ce{11 H+} \\ &\text{right side} &\quad 3\cdot(+3) + 2\cdot(+3) &= +15 &\quad &|\text{add}~\ce{11 H+ and 7 O} \\ \hline &\text{difference} &\quad &= +11 \end{align}

But this doesn't align with the solution for this problem, which is:

$$\ce{6Fe^2+(aq) + Cr2O7^2-(aq) + 14H+(aq) -> 6Fe^3+(aq) + 2Cr^3+(aq) + 7H2O}$$

Why am I getting three iron atoms and not six?

• Now, why are six iron atoms added?
– Nora
Commented Feb 22 at 14:19
• There is this magical formula 2 x (6-3)=6 x (3-2) // By other words, the sum of changes of oxidation numbers must be zero, to honor the charge conservation law. Commented Feb 22 at 15:39
• In the question, you have forgotten there are 2 Cr in Cr2O7^2-. Therefore 1 Cr2O7^2-. is able to oxidize 6 Fe^2+ and not just 3. Commented Feb 22 at 16:20
• Hmm, alright. Thank you. I thought you were being sarcastic when you said "magic formula". My bad. Thanks =p
– Nora
Commented Feb 22 at 16:24
• I use it often in general form m.n=n.m. As when chemical equations are concerned, many people forget basics of arithmetics of small integers and the commutative law. If they forget it is chemistry, they can see it is simple. so 2 times 3 = 6 times 1. Commented Feb 22 at 16:26

On the first line, there was a title : "The problem". The second line is : "Balance the chemical equation using oxidation numbers". The third line is : "What I have tried". The fourth line is the equation between $$\ce{Fe^{2+}}$$ and $$\ce{Cr2O7^{2-}}$$, which is correct, although no stoichiometric coefficients are indicated. Up to here all is right. Nothing to object. The main problem here is to define these coefficients. This is where you made a mistake.
The next two lines start with "Electron transition", follow by the dichromate reduction. Here you correctly find that $$6$$ electrons are needed when reducing $$1$$ $$\ce{Cr2O7^{2-}}$$ ion to $$2$$ $$\ce{Cr^{3+}}$$, although the formula of the chromium ions has a $$+3$$ charge, which was forgotten. Well. This typo has no consequence.
What is more serious is the following line, where you correctly state that the oxidation of $$\ce{Fe^{2+}}$$ to $$\ce{Fe^{3+}}$$ produces $$1$$ electron. $$1$$ electron par iron ion is correct. But just after this equation you multiply this equation by $$3$$. Strange decision. Why by $$3$$ ? There is no reason for multiplying by $$3$$ the equation $$\ce{Fe^{2+} -> Fe^{3+} + e-}$$ If you multiply this equation by $$3$$, you produce $$3$$ electrons. And these $$3$$ electrons cannot do anything ! To reduce one $$\ce{Cr2O7^{2-}}$$ ion, $$3$$ electrons are not useful. Instead of $$3$$, the ion $$\ce{Cr2O7^{2-}}$$ requires $$6$$ electrons. Six ! Not three ! The final equation must bring $$6$$ electrons ; so it must contain $$6$$ $$\ce{Fe^{2+}}$$ ions on the left-hand side.
This is the origin of your mistake. Once you have understood the origin this number $$6$$, I am sure you will find the rest of the problem, and obtain the final equation.