Answer :
Answer:
[tex]K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)[/tex]
Explanation:
The products of this reaction are given by:
[tex]K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)[/tex]
Firstly, dichromate anion becomes chromium(III) cation, let's write this change:
[tex]Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)[/tex]
The following steps should be taken:
- balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:
[tex]Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)[/tex]
- balance oxygen atoms by adding 7 water molecules on the right:
[tex]Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)[/tex]
- balance the hydrogen atoms by adding 14 protons on the left:
[tex]Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)[/tex]
- balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):
[tex]Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)[/tex]
Similarly, tin(II) cation becomes tin(IV) cation:
[tex]Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-[/tex]
Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:
[tex]Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)[/tex]
[tex]3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-[/tex]
Add them together:
[tex]Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)[/tex]
Adding the ions spectators:
[tex]K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)[/tex]