Answer:
[tex]Cr_2O_7^{2-}(aq)+2Cu^+(aq)\rightarrow Cu_2Cr_2O_7(s)[/tex]
Explanation:
Hello!
In this case, when writing net ionic equations, we first need to identify the complete molecular equation; thus for potassium dichromate reacting with copper (I) bromide, we have:
[tex]K_2Cr_2O_7(aq)+2CuBr(aq)\rightarrow Cu_2Cr_2O_7(s)+2KBr(aq)[/tex]
Thus, since all the species are aqueous except copper (II) dichromate, we can write the complete ionic equation by ionizing the aqueous ones:
[tex]2K^++Cr_2O_7^{2-}+2Cu^++2Br^-\rightarrow Cu_2Cr_2O_7(s)+2K^++2Br^-[/tex]
Finally, for the net ionic equation, we need to cancel out the spectator ions as they are present at both reactants and products sides, thus, we obtain:
[tex]Cr_2O_7^{2-}(aq)+2Cu^+(aq)\rightarrow Cu_2Cr_2O_7(s)[/tex]
Best regards!
