Question

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

Answer 1

`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)`

Step-by-step 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)`