Why does balancing the reduction of dichromate(VI) to chromium(III) require four water molecules among the products?

Question

Problem

Balance the following chemical equation using oxidation numbers:

$$\ce{Cr2O7^2–(aq) + HNO2(aq) –> Cr^3+(aq) + NO3–(aq)}$$

Solution (in Swedish)

$$\ce{\overset{+VI}{Cr}_2O7^2–(aq) + H\overset{+III}{N}O2(aq) –> \overset{+III}{Cr}^3+(aq) + \overset{+V}{N}O3–(aq)}$$

\begin{align} \text{Elektronövergång:}\quad \ce{\overset{+VI}{Cr}_2 + 6 e- &-> 2 \overset{+III}{Cr}} &\quad &|\times 1 \\ \ce{\overset{+III}{N} &-> \overset{+V}{N} + 2 e-} &\quad &|\times 3 \end{align}

$$\ce{Cr2O7^2–(aq) + 3 HNO2(aq) –> 2 Cr^3+(aq) + 3 NO3–(aq)}$$

\begin{align} \text{Laddningar:}\quad &\text{vänsterled} &\quad &= -2 &\quad &|\text{lägg till}~\ce{5 H+} \\ &\text{högerled} &\quad 2\cdot(+3) + 3\cdot(-1) &= +3 &\quad &|\text{lägg till}~\ce{4 H2O} \\ \hline &\text{skillnad} &\quad &= +5 \end{align}

$$\ce{Cr2O7^2–(aq) + 3 HNO2(aq) + 5 H+(aq) –> 2 Cr^3+(aq) + 3 NO3–(aq) + 4 H2O}$$

Question

I can understand everything about the solution, except for the added $\ce{4 H2O}$ on the right side of the equation. Why is it that $\ce{4 H2O}$ is added to the right side and not $\ce{2.5 H2O}?$ How come the equation is balanced despite the oxygen atoms not being equal on both sides?

Answer 1

$$\ce{Cr2O7^2–(aq) + 3 HNO2(aq) + 5 H+(aq) -> 2 Cr^3+(aq) + 3 NO3–(aq) + 4 H2O} $$ The left side has $1\cdot7 + 3\cdot2 = 13$ oxygen atoms and $3\cdot1 + 5\cdot1 = 8$ hydrogen atoms. The right side has $3\cdot3 + 4\cdot1 = 13$ oxygen atoms and $4\cdot2 = 8$ hydrogen atoms. It is not clear to my why you would expect 2.5 water molecules instead of 4. You can't just count the $\ce{5 H+}$, you have to consider that $\ce{HNO2}$ contains hydrogen as well.