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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}} &\quad &|\times 1 \\ \ce{\overset{+II}{Fe} &-> \overset{+III}{Fe} + 1 e-} &\quad &|\times 3 \end{align}\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?

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}} &\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?

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?

Used \ce{} two times.
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Metal Storm
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The problem:

Balance the following chemical equation using oxidation numbers:

$Fe^{2+} + Cr_2{O_7}^{2-} \rightarrow Fe^{3+} + Cr^{3+}$$$ \ce{Fe^2+ + Cr_2O_7^2- -> Fe^3+ + Cr^3+} $$

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}} &\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:

$6Fe^{2+}(aq) + Cr_2{O_7}^{2-}(aq) + 14H^+(aq) → 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O$$$ \ce{6Fe^2+(aq) + Cr2O7^2-(aq) + 14H+(aq) -> 6Fe^3+(aq) + 2Cr^3+(aq) + 7H2O} $$

Why am I getting three iron atoms and not six?

The problem:

Balance the following chemical equation using oxidation numbers:

$Fe^{2+} + Cr_2{O_7}^{2-} \rightarrow Fe^{3+} + Cr^{3+}$

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}} &\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:

$6Fe^{2+}(aq) + Cr_2{O_7}^{2-}(aq) + 14H^+(aq) → 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O$

Why am I getting three iron atoms and not six?

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}} &\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?

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Nora
Nora
  • 133
  • 5

The problem:

Balance the following chemical equation using oxidation numbers:

$Fe^{2+} + Cr_2{O_7}^{2-} \rightarrow Fe^{3+} + Cr^{3+}$

What I have tried:

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

\begin{align} \text{Electron transition:}\quad \ce{\overset{+VI}{Cr}_2 + 6 e- &-> 2 \overset{+III}{Cr}} &\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:

$6Fe^{2+}(aq) + Cr_2{O_7}^{2-}(aq) + 14H^+(aq) → 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O$

Why am I getting three iron atoms and not six?

The problem:

Balance the following chemical equation using oxidation numbers:

$Fe^{2+} + Cr_2{O_7}^{2-} \rightarrow Fe^{3+} + Cr^{3+}$

What I have tried:

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

\begin{align} \text{Electron transition:}\quad \ce{\overset{+VI}{Cr}_2 + 6 e- &-> 2 \overset{+III}{Cr}} &\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:

$6Fe^{2+}(aq) + Cr_2{O_7}^{2-}(aq) + 14H^+(aq) → 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O$

Why am I getting three iron atoms and not six?

The problem:

Balance the following chemical equation using oxidation numbers:

$Fe^{2+} + Cr_2{O_7}^{2-} \rightarrow Fe^{3+} + Cr^{3+}$

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}} &\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:

$6Fe^{2+}(aq) + Cr_2{O_7}^{2-}(aq) + 14H^+(aq) → 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O$

Why am I getting three iron atoms and not six?

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Nora
Nora
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