where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2 M x V1 = 0.1 M x .5 L

<span>V1 = 0.025 L or 25 mL of the 2 M KCl solution is needed</span>

3 0

3 years ago

(ii) The Cu2+ ions are attracted to the negative electrode, where they are reduced to produce copper atoms.

allochka39001 [22]

Electrolyte for the electrolysis is copper(II) sulfate
CuSO₄ = Cu²⁺ + SO₄²⁻

the reaction at the negative electrode
(cathode -) Cu²⁺ + 2e⁻ = Cu⁰

and...
(anode +) 2H₂O - 4e⁻ = O₂ + 4H⁺

5 0

3 years ago

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

Mumz [18]

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:

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

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