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3 years ago

Acidified solutions of sodium dichromate, K 2 Cr 2 O 7 , and copper (I) bromide, CuBr, are mixed.

Chemistry

1 answer:

Nina [5.8K]3 years ago

6 0

Answer:

$Cr_2O_7^{2-}(aq)+2Cu^+(aq)\rightarrow Cu_2Cr_2O_7(s)$

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:

$K_2Cr_2O_7(aq)+2CuBr(aq)\rightarrow Cu_2Cr_2O_7(s)+2KBr(aq)$

Thus, since all the species are aqueous except copper (II) dichromate, we can write the complete ionic equation by ionizing the aqueous ones:

$2K^++Cr_2O_7^{2-}+2Cu^++2Br^-\rightarrow Cu_2Cr_2O_7(s)+2K^++2Br^-$

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:

$Cr_2O_7^{2-}(aq)+2Cu^+(aq)\rightarrow Cu_2Cr_2O_7(s)$

Best regards!

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