Question

In: Chemistry

c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) Is this a redox reaction? (yes or...

c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Is this a redox reaction? (yes or no) _____________

If yes:

i. what is the oxidizing agent? ___________________

ii. what is the reducing agent? ___________________

iii. what species is being oxidized? __________________

iv. what species is being reduced? __________________

Expert Solution

Answer :-

This reaction is redox reaction.

oxidizing agent :- Hydrogen (H)

reducing agent :- Zinc (Zn)

species is being oxidized :- Zinc (Zn)

species is being reduced :- Hydrogen (H)

.

*************

Explanation :-

The oxidation number of Hydrogen is decreased from +1 to 0. i.e. Hydrogen undergoes reduction and act as oxidising agent.

The oxidation number of Zinc is increased from 0 to +2. i.e. Zinc undergoes oxidation and act as reducing agent.

**********

Note :-

Oxidation :-

The loss of electrons is called oxidation.

The oxidation number of oxidized species increases.

Reduction :-

The gain of electrons is called reduction.

The oxidation number of oxidized species decreases.

Redox reaction :-

The reaction in simultaneous oxidation and reduction take place is called redox reaction.

Oxidising Agent (Oxidant)	Reducing Agent (Reductant)
Causes oxidation	Causes reduction
Accepts electron(s)	Donates electron(s)
Itself undergoes reduction	Itself undergoes oxidation
Undergoes decrease in oxidation number	Undergoes increase in oxidation number

.

*******************

queen_honey_blossom answered 2 years ago

Concerning the reaction: 2 HCl(aq) + Zn(s)  H2(g) + ZnCl2(aq) A piece of Zn with...

Concerning the reaction: 2 HCl(aq) + Zn(s)  H2(g) + ZnCl2(aq) A piece of Zn with a mass of 3.2 g is placed in 233 mL of 0.19 M HCl (aq). What mass in grams of H2(g) is produced in the cases of: a) 100. % yield b) 63.0 % yield

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) How many...

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) How many liters of H2 would be formed at 636 mm Hg and 17 ∘C if 27.0 g of zinc was allowed to react? How many grams of zinc would you start with if you wanted to prepare 6.85 L of H2 at 284 mm Hg and 28.5 ∘C?

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl : Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) 1)How...

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl : Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) 1)How many liters of H2 would be formed at 592 mm Hg and 17 ∘C if 25.5 g of zinc was allowed to react? 2)How many grams of zinc would you start with if you wanted to prepare 4.60 L of H2 at 294 mm Hg and 34.5 ∘C ?

Zn(s) + HCL(aq) ----> Zn 2+(aq) +Cl-(aq) + H2(g) + heat. Balance the equation and: a)calculate...

Zn(s) + HCL(aq) ----> Zn 2+(aq) +Cl-(aq) + H2(g) + heat. Balance the equation and: a)calculate the number of moles of hydrogen gas that can be produced from 1.0g of zinc (6.38g/mol) pellets, assuming the reaction completes. b)Calculate the final temperature of the solution of 10.0 mL of 1.0 M aqueous HCl at an initial temperature of 20.7 degrees Celsius was used to consume the zinc pellets. Use deltaHrxn= -80.7 kJ/mol and assume that the hydrogen gas doesn't affect the...

Consider the reaction: Zn(s) + 2 H+(aq) = Zn2+(aq) + H2(g) At 25C, calculate: a) ∆G˚...

Consider the reaction: Zn(s) + 2 H+(aq) = Zn2+(aq) + H2(g) At 25C, calculate: a) ∆G˚ for the reaction, given that: ∆Gf Zn(s) = 0, ∆Gf(H+) = 0, ∆Gf(H2) = 0, ∆Gf(Zn2+) = -147.1 kj/mol b) ∆G, when P(h2) = 750 mmHg, [Zn2+ aq] = 0.10 M, [H+] = 1.0 x 10^-4 M c) The pH when ∆G - -100 kJ, P(h2) = 0.922 atm, [Zn2+] = 0.200 M and the mass of Zn is 155 g.

Reaction of HCl (aq) and Mg(s) to form MgCl2(aq) and H2(g) Given: Molarity of HCL :...

Reaction of HCl (aq) and Mg(s) to form MgCl2(aq) and H2(g) Given: Molarity of HCL : 3.00M volume of HCL: 20.0mL mass of Mg: 0.036g volume of gas before placing in equalization chamber: 37.0mL volume of gas after placing in equalization chamber: 37.5 mL barometric pressure of the room: 736.4 mmhg temperature of the room: 18.5 C ------------------------------------------------ Calculate: mole of Mg reacted: mole of H2(g) formed: vapor pressure of vater: pressure of H2 (Daltons Law): pressure of H2 in...

Zinc metal reacts with hydrochloric acid according to the following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When 0.107 g...

Zinc metal reacts with hydrochloric acid according to the following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When 0.107 g of Zn(s) is combined with enough HCl to make 54.9 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.0 ∘C to 24.0 ∘C. Find ΔHrxn for this reaction as written. (Use 1.0 g/mL for the density of the solution and 4.18 J/g⋅∘C as the specific heat capacity.)

Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh...

Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh out a 0.250 gram piece of metal and combine it with 72.9 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 42.29 g/mol, and you measure that the reaction absorbed 155 J of heat, what is the enthalpy of this reaction in kJ per mole of limiting reactant? Enter your answer numerically to three significant figures...

ZnCl2 (aq) + Al (s) ---- Zn(s) + AlCl3 (aq) [Al3+] = 0.8 M [Zn2+] =...

ZnCl2 (aq) + Al (s) ---- Zn(s) + AlCl3 (aq) [Al3+] = 0.8 M [Zn2+] = 1.3M Zn2+(aq) +2e- ---Zn(s) -0.76V Al3+(aq) + 3e- --- Al (s) -1.66V Balance the equation above. Give the equation for the reaction quotient(Q) and solve for it. Are there units for (Q)? Solve for the standard potential for the cell (EO cell) at 295K. First give equation. Solve the Nernst equation for this cell at 295K.

a) A standard ZnCl2 solution is prepared by dissolving 0.6328 g of Zn in an HCl...

a) A standard ZnCl2 solution is prepared by dissolving 0.6328 g of Zn in an HCl solution and diluting to volume in a 1.00 L volumetric flask. An EDTA solution is standardized by titrating a 10.00 mL aliquot of the ZnCl2 solution, which requires 10.84 mL of EDTA solution to reach the end point. Determine the concentration of the EDTA solution. b) A 1.4927 g sample of powdered milk is dissolved and the solution titrated with the EDTA solution prepared...