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Leni [432]
3 years ago
12

Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl​

Chemistry
1 answer:
Mumz [18]3 years ago
7 0

Answer:

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)

Explanation:

The products of this reaction are given by:

K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)

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:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)

  • balance oxygen atoms by adding 7 water molecules on the right:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the hydrogen atoms by adding 14 protons on the left:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • 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):

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

Similarly, tin(II) cation becomes tin(IV) cation:

Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-

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:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-

Add them together:

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)

Adding the ions spectators:

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)

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