Question

the two half-reactions are given below.
\\(14\text{h}^+ + \text{cr}_2\text{o}_7^{2-} + 6\text{e}^- \
ightarrow 2\text{cr}^{3+} + 7\text{h}_2\text{o}\\)
\\(2\text{cl}^- \
ightarrow 2\text{e}^- + \text{cl}_2\\)
what is the coefficient for the \\(\text{cl}_2\\) in the balanced net ionic equation?
hint: make sure the electrons cancel when you combine the two half-reactions.

Explanation:

Step1: Balance electrons in half - reactions

The first half - reaction: \(14H^{+}+Cr_{2}O_{7}^{2 -}+6e^{-}
ightarrow2Cr^{3 +}+7H_{2}O\) (has 6 electrons gained)
The second half - reaction: \(2Cl^{-}
ightarrow2e^{-}+Cl_{2}\) (has 2 electrons lost per reaction). To balance the number of electrons, we need to multiply the second half - reaction by 3 so that the number of electrons lost equals the number of electrons gained.
After multiplying the second half - reaction by 3: \(3\times(2Cl^{-}
ightarrow2e^{-}+Cl_{2})\) gives \(6Cl^{-}
ightarrow6e^{-}+3Cl_{2}\)

Step2: Combine the two half - reactions

Now we add the two half - reactions together. The first half - reaction is \(14H^{+}+Cr_{2}O_{7}^{2 -}+6e^{-}
ightarrow2Cr^{3 +}+7H_{2}O\) and the modified second half - reaction is \(6Cl^{-}
ightarrow6e^{-}+3Cl_{2}\)
When we add them, the electrons (\(6e^{-}\)) cancel out. The combined reaction is \(14H^{+}+Cr_{2}O_{7}^{2 -}+6Cl^{-}
ightarrow2Cr^{3 +}+7H_{2}O + 3Cl_{2}\)

Answer:

3