Answer:
[tex]Cr_2O_7^{2-}(aq)+7H_2O(l)+ 6e^-\rightarrow 2Cr^{3+}\\\\Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+1e^-[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to write out the balanced half reactions by considering that chromium is being reduced and iron oxidized from 6+ to 3+ and 2+ to 3+ respectively, in such a way, we have:
[tex]Cr_2O_7^{2-}(aq)\rightarrow 2Cr^{3+}\\\\Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)[/tex]
Now, we insert the appropriate amount of water molecules and the carried electrons in order to balance both oxygen and consequently hydrogen:
[tex]Cr_2O_7^{2-}(aq)+7H_2O(l)+ 6e^-\rightarrow 2Cr^{3+}\\\\Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+1e^-[/tex]
Which is further transformed to the equation given in the problem due to the electron balance.
Regards!
