# How could you balance this redox equation involving Cr2O7 2- and H2O

Hi I know that balancing redox equations you first work out the separate half equations first and then combine them however I'm not sure how to balance this one?

$$\ce{Cr_2O_7^{2-} + H^+(aq) + e^- \rightarrow Cr^{3+}(aq) + H_2O(l)}$$

I've arried at $\ce{Cr_2O_7^{2-} + 3e^- \rightarrow Cr^{3+}}$ however for $\ce{H^+ \rightarrow H_2O}$ I'm not sure what to add because the oxidation for state of hydrogen is the same?

• Why would you add 3e- in the first place? Also, welcome to Chem.SE, and pay attention to the formatting. – Ivan Neretin Aug 3 '16 at 6:14
• @IvanNeretin Hi I found the oxidation number for Cr and it was +6 on the left hand side so I thought that you would need 3e- to balance the difference in oxidation states? – cz88 Aug 3 '16 at 7:23
• That would be right if not for the fact that you have two of them Cr. – Ivan Neretin Aug 3 '16 at 7:33
• But wouldn't one Cr at the products be +6 so that both of them would give me +12? so Cr would be +6? @IvanNeretin – cz88 Aug 3 '16 at 11:34
• No, they both would turn from +6 to +3. – Ivan Neretin Aug 3 '16 at 11:43

To balance a redox reaction (in acid),

1. First balance the atom being reduced or oxidized:

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

1. Then balance the oxygens by adding $\ce{H_2O}$:

$$\ce{Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7 H_2O}$$

1. Then balance the hydrogen by adding $\ce{H^+}$:

$$\ce{Cr_2O_7^{2-} + 14 H^+ \rightarrow 2Cr^{3+} + 7 H_2O}$$

1. Finally, balance the total charge by adding $\ce{e^-}$:

$$\ce{Cr_2O_7^{2-} + 14 H^+ + 6e^- \rightarrow 2Cr^{3+} + 7 H_2O}$$