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What is the correct cell notation for the reaction 3Zn2 plus aq plus 2Als equals 2Al3 plus aq plus 3Zn s?
Wiki User
∙ 13y ago
Al(s) | Al3+(aq) Zn2+(aq) | Zn(s)
Wiki User
∙ 13y ago
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What is the balanced equation for the reaction of phosphoric acid and aluminum?
3Al3 + 6H3PO4 -> 9H2 + 2Al3(PO4)3
What is the balanced equation for the reaction of magnesium metal with aluminium ions?
3Mg (s) + 2Al3+ (aq) ---> 3Mg2+ (aq) + 2Al (s)
What is the net ionic equation for hydrochloric acid and aluminum?
6H3O+ + 2Al --> 2Al3+ + 3H2 + 6H2O[Chloride ions (Cl -) are left out of equation, don't take part in reaction: 'tribune' ions].
Will be there a reaction if aluminum is dipped into zinc nitrate solution?
Yes, Al is more reactive than Zn, so: 2Al + 3Zn2+ --> 2Al3+ + 3Zn (Nitrate ions are tribuned out of this reaction)
What is the difference between a total ionic equation and a net ionic equation?
The net ionic equation has only the species involved in the chemical reaction.
What is the chemical equation for aluminum oxide plus carbon aluminum plus carbon dioxide?
Aluminum oxide does not react with water! Not under normal conditions anyway. That's why aluminum metal doesn't corrode -- but a thin aluminum oxide layer builds up and protect the outer surface.That being said, it can react with water in acidic conditions to produce Al3+ ions.Al2O3(s) + 6H(aq) --> 2Al3+(aq) + 3H2O(l)
An ionic bond is attracted between what?
.... between one or more postive (metal) ions and one or more negative ions like chloride, sulfate, nitrate, carbonate etc. Example: 2Al3+ + 3SO42- --> Al2(SO4)3
What happens when aluminum is added to water and copper chloride?
The aluminum metal appears to be turning into copper, but it is actually just removing the metallic copper from its compound state. The Aluminum is oxidized and loses e-, becoming Al 3+ and the copper ions are reduced (they accept those e-) to form solid copper precipitate. The aluminum ions and chloride ions remain in the solution. The reaction will only occur in water and occurs because the transfer of electrons from the aluminum to the copper results in a more stable system.
What is the ionic bonds is the attraction between?
.... between one or more postive (metal) ions and one or more negative ions like chloride, sulfate, nitrate, carbonate etc. Example: 2Al3+ + 3SO42- --> Al2(SO4)3
Is aluminum oxide an oxide?
Aluminum*)oxide is an amphoteric compound by being both 1. acidic and 2. alkaline1. acidicIt will dissolve in hydroxide (an alkaline solution):Al2O3 + 2OH- + 3H2O --> 2 Al(OH)4- aluminate anion2. alkalineIt will dissolve in acidic H+ solution:Al2O3 + 6H+ --> 2Al3+ + 3H2OThe amphoteric, neutral formula can be written as both Al2O3.(H2O)3 and: 2Al(OH)3Added(The same answer, but this time written without ions in solvent):Aluminium*)oxide is amphoteric:Al2O3(s) can react with a base:Al2O3(s) + 2NaOH(aq) + H2O(l) --> 2NaAl(OH)4(aq)Aluminium oxide + sodium hydroxide + water --> sodium aluminateAl2O3 can react with an acid:Al2O3(s) + 6HCl(aq) --> 2AlCl3(aq) + 3H2O(l)Aluminium oxide + hydrochloric acid --> aluminum chloride + water*) The USA spells Al as aluminum; British Isles as aluminum. Both are correct.
Which is an important step in the alternate method for balancing equations in redox reactions?
Follow these 6 stepsNOTE: This is NOT the same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY!These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction:CuS(s) + NO3-(aq) ---> Cu2+(aq) + SO42-(aq) + NO(g)Step 1:Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product.Here is the unbalanced half-reaction involving CuS:CuS(s) ---> Cu2+(aq) + SO42-(aq)And the unbalanced half-reaction for NO3- is:NO3-(aq) --> NO(g)Step 2:Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction.In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here.Step 3:Balance oxygen by adding H2O to one side of each half-reaction.CuS + 4 H2O ---> Cu2+ + SO42-NO3- --> NO + 2 H2OStep 4:Balance hydrogen atoms. This is done differently for acidic versus basic solutions.For acidic solutions: Add H3O+ to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H2O to the other side.For basic solutions: add H2O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH- to the other side.Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have:CuS + 12 H2O ---> Cu2+ + SO42- + 8 H3O+NO3- + 4 H3O+ --> NO + 6 H2OStep 5:Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.Oxidation: CuS + 12 H2O ---> Cu2+ + SO42- + 8 H3O+ + 8 e-Reduction: NO3- + 4 H3O+ + 3 e- --> NO + 6 H2OStep 6:Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H3O+, H2O, or OH- appears on both sides of the final equation, cancel out the duplication also.Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed).3 CuS + 36 H2O ---> 3 Cu2+ + 3 SO42- + 24 H3O+ + 24 e-8 NO3- + 32 H3O+ + 24 e- ---> 8 NO + 48 H2OAdding these two together gives the following equation:3 CuS + 36 H2O + 8 NO3- + 8 H3O+ ---> 3 Cu2+ + 3 SO42- + 8 NO + 48 H2OStep 7:Finally balancing both sides for excess of H2O(On each side -36) This gives you the following overall balanced equation at last:3 CuS(s) + 8 NO3-(aq) + 8 H3O+(aq) ---> 3 Cu2+(aq) + 3 SO42-(aq) + 8 NO(g) + 12 H2O(l)
What scientific conclusion can be made about combining Copper Chloride and Aluminum Foil?
The two are highly reactive when mixed in water. If you add aluminum to a water solution of copper chloride, the aluminum will be combined as aluminum chloride gas, releasing the copper into the water. The mixture will turn very hot, bubble up, and start to smoke, and the color will change from blue to dark red. ========================= The first sentence is correct, the two are highly reactive, but from there on, there are problems... There's no such thing as aluminum chloride gas. There is a gas given off, but it is hydrogen gas. The "smoke" is actually tiny droplets of water condensed from water vapor. It gets that hot. The red color is due to the formation of copper metal a Cu2+ ions are reduced to the metal as aluminum metal is oxidized. Aluminum metal has a thin coating of aluminum oxide, Al2O3, covering the surface. Even freshly scraped aluminum metal will quickly reform the passivating layer of aluminum oxide. So in many solutions, the layer of Al2O3 prevents any aluminum metal from reacting. If aluminum metal is placed in a solution of copper(II) sulfate, you will not get a reaction. No copper metal will form on the aluminum. But when placed in copper(II) chloride, aluminum will give a vigorous reaction with a lot of heat and hydrogen gas being given off. The reason is the chloride ion. In the presence of chloride ion, the Al2O3 layer dissolves forming the AlCl4^- in solution and exposing a fresh layer of aluminum metal. The aluminum metal will reduce copper(II) ions to copper metal, AND the aluminum metal will react with water very much like an alkali metal reacts with water, vigorously, with a lot of heat given off, and with the formation of hydrogen gas. Al2O3 + 8Cl- + 3H2O --> 2AlCl4^- + 6OH- Al(s) + OH- + 2H2O --> Al(OH)3(s) + H2(g) 2Al(s) + 3Cu2+ --> 2Al3+ + 3Cu(s)
