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3 years ago

Metal rings can be coated with a layer of copper using electricity.

Chemistry

1 answer:

Eduardwww [97]3 years ago

5 0

<u>First of all, what is electrolysis?</u>

Electrolysis is the process of breaking down ionic substances using direct current.

<u>Important points about electrolysis </u>

→ Ionic substances contain particles called ions.

→ Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions. The ions must be free to move for electrolysis to occur and it can happen by melting or dissolving an ionic substance in water.

→ Positively charged ions move to the negative electrode. They receive electrons and are <em>reduced</em>. The positive ions move towards the negative electrode because they want to cancel each other out.

→ Negatively charged ions move to the positive electrode. They lose electrons and are <em>oxidised</em>. The substance that is broken down is called the electrolyte <em>(an electrolyte is just a liquid or solution that can conduct electricity)</em> . The negative ions move towards the positive electrode because they want to cancel each other out.

<h3>Cathode = Negative electrode</h3><h3>Anode = Positive electrode</h3>

Metal ions form at the cathode and non-metal ions form at the anode

How I remember if an element is <em>oxidised</em> or <em>reduced</em> is by remembering OIL RIG

OIL = Oxidation is Loss (of electrons)

RIG = Reduction is Gain (of electrons)

<h2><em><u>The answer to your question</u></em></h2>

1) The first step would be to clean the metal ring and sand it down because when the metal atoms from the electrolyte are deposited onto the ring, they will form a weak bond and they may simply 'fall' off. Also this could affect conductivity and the whole experiment. The more things you do accurately now, the more accurate your result will be.

2) You want to put the solution you are given in to the tank your going to be using.

3) This is basically the main part, you want to set up the circuit, I have attached a diagram at the bottom to show you the circuit. The copper rod will be the anode and the metal ring will be a cathode (ignore the elements).

4) Now turn on the circuit and you will start to see the solution spilt with the the solution now being split some going to the anode and some going the cathode.

5) Then a thin layer should form on the electrode.

Hope this helps :)

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For the following reactions, predict the products and write the balanced formula equation, complete ionic equation, and net ioni

stealth61 [152]

Answer:

Explanation:

To predict the products of these reactions we need to know the kind of reactions. All these reactions are double replacement reaction. In these kinds of reactions, the products will be the result of exchanging ions in the reactants. So, the first step is to identify the ions.

For the reaction, we have Hg2(NO3)2 and CuSO4. We have the ions Hg+1, NO3-1, Cu+2 and SO4-2

The way to make this exchange is putting together positive in one species with the negative of the other species. Following that rule we have

$Hg^{+1} - - - (SO_{4})^{-2}[/text]
the oxidation number will tell you the subscript for each species in the compound. In this case, is Hg2(SO4) [tex]Cu^{+2} - - - (NO_{3})^{-1} - - -> Cu(NO_{3})_{2} [/text] 
So, the products for this reaction will be
 [tex]Hg_{2} (NO_{3})_{2}(aq) + CuSO_{4}(aq) --> Hg_{2}SO_{4} + Cu(NO_{3})_{2}[/text]

After this, we proceed to balance the equation. For this, we check that we have the same number of each element on both sides of the equation. In this case, we can see that we have the same number, so the equation is balanced. Finally, we check the rules of solubility to see if the species are soluble in water or not. In this case sulfates area always soluble except for mercury so Hg2(SO4) precipitates in the solution (pre). Nitrates are always soluble so Cu(NO3)2 is soluble (aq) 
[tex] Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq) - -> Hg_{2}SO_{4} (pre) + Cu(NO_{3})_{2}(aq)$

The complete ionic equation allows to show which of the reactants or products exist primarily as ions. For this reaction this will be:

$2Hg^{+1}(aq) + 2(NO_{3})^{-1}(aq) + (SO_{4})^{-2}(aq) + Cu^{+2}(aq) --> Hg_{2}SO_{4} (pre)+ Cu^{+2}(aq) + (NO_{3})^{-1}(aq) [/text]

To get net ionic equation we take away the ions that did not participate in the reactions. In other words the ones that are the same on both sides in the equation. In this case we see that [tex] Cu^{+2}(aq) and (NO_{3})^{-1}(aq) [/text] are the same on both sides so those ions are not include in the net ionic equation. This is:
[tex] 2Hg^{+1}(aq) + (SO_{4})^{-2}(aq) --> Hg_{2}SO_{4} (pre) [/text]

B [tex] Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq)$

ions (1) $Ni^{+2} and (NO_{3})^{-1}$

ions (2) $Ca^{+2} and Cl^{-1}$

Exchanging

$Ni^{+2} ---- Cl^{-1} --> NiCl_{2}$

$Ca^{+2} --- (NO_{3})^{-1} --> Ca(NO_{3})_{2}$

Products

$Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq) --> NiCl_{2} + Ca(NO_{3})_{2}$

The equation is already balanced

Chlorides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NiCl2 is soluble (aq)

Nitrates are always soluble. Ca(NO3)2 is soluble (aq)

Since both compounds are soluble, we can say that there is not reaction.

Complete ionic equation

$Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq) - - > Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq)$

Net ionic equation:

The ions in both sides of the equation are the same so all of them are cancelled and we cannot get a net ionic equation this explains why there is no reaction in this case.

C $K_{2}CO_{3}(aq) + MgI_{2}(aq)$

Ions(1) $K^{+1} and (CO_{3})^{-2}$

Ions(2) $Mg^{+2} and l^{-1}$

Exchanging

$K^{+1} --- l^{-1} - - > KI$

$Mg^{+2} --- (CO_{3})^{-2} - - > Ca(CO_{3})$

Products

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> Kl + MgCO_{3}$

The equation is not balanced

Balance equation is

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> 2Kl (aq) + MgCO_{3} (pre)$

iodides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. KI is soluble (aq)

carbonates are always insoluble except group 1 cations. MgCO3 is insoluble (pre)

complete ionic equation

$2K^{+1}(aq) + (CO_{3})^{-2}(aq) + Mg^{+2}(aq) + 2l^{-1}(aq) - - > MgCO_{3} (pre) + 2K^{+1}(aq) + 2l^{-1}(aq)$

Net ionic equation

$(CO_{3})^{-2}(aq) + Mg^{+2}(aq) - - > MgCO_{3} (pre)$

D $Na_{2}CrO_{4}(aq) + AlBr_{3}(aq)$

Ions(1) $Na^{+1} and (CrO_{4})^{-2}$

Ions(2) $Al^{+3} and Br^{-1}$

Exchanging

$Na^{+1} ---- Br^{-1} - -> NaBr$

$Al^{+3} --- (CrO_{4})^{-2} - -> Al_{2}(CrO_{4})_{3}$

Products

$Na_{2}CrO_{4}(aq) + AlBr_{3}(aq) - -> NaBr + Al_{2}(CrO_{4})_{3}$

The equation is not balanced

Balance equation is

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr + Al_{2}(CrO_{4})_{3}$

bromides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NaBr is soluble (aq)

chromates are always insoluble except group 1 cations. Al2(CrO4)3 is insoluble (pre)

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr(aq) + Al_{2}(CrO_{4})_{3}(pre)$

Complete ionic equation

$6Na^{+1}(aq) + 3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) + 6Br^{-1}(aq) - -> Al_{2}(CrO_{4})_{3}(pre) +6Br^{-1}(aq) + 6Na^{+1}(aq)$

Net ionic equation

$3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) - -> Al_{2}(CrO_{4})_{3}(pre)$

6 0

3 years ago

PLS HELP ME WITH THIS ILL GIVE BRAINLIEST

anyanavicka [17]

Answer:

1 qualiative physical

2 qualiative chemical

3 quantitative physcial

4 quantiative chemical

5qualiative physcial

6quatiative chemcial

7quanatiative physcial

8 qualiative chemical

9quatliate physcial

6 0

2 years ago

Materials Visible Uniform Not Visible Non- Uniform State of Matter 1. Sugar and water 2. Oil and water 3. Vinegar and water 4. S

salantis [7]

Answer:

1. Uniform, non visible

2. Visible, non uniform

3. Uniform, non visible

4. Visible, non uniform

5. Visible, non uniform

Explanation:

A heterogeneous mixture is simply any mixture that is not uniform in composition — it's a non-uniform mixture of smaller constituent parts. By contrast, a mixture that is uniform in composition is a homogenous mixture

visible state of matter is when you can easily difference two components of a mixture

non visible is when the state of the matter in the substance is not visible to the eye, it can't be differentiated

4 0

2 years ago

Read 2 more answers

A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution? Us

drek231 [11]

Answer:6.0M

Explanation:see attached photo