Ask question

Ask question

All categories

3 years ago

11

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

e flow of electrons through the external circuit in this cell is from the

1 answer:

soldi70 [24.7K]3 years ago

3 0

Answer: from the Zn anode to the Cu cathode

Justification:

1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)

2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).

3) Then, you can already tell that electrons go from Zn to Cu.

4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.

So you get that the electrons flow from the anode (Zn) to the cathode (Cu).

Always oxidation occurs at the anode, and reduction occurs at the cathode.

You might be interested in

Complete the equation for the dissociation of Na2CO3(aq).Omit water from the equation because it is understood to be present.

ozzi

Answer:

Na2CO3 <==> 2Na^+ + CO3^2-

Explanation:

6 0

3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

nikdorinn [45]

Answer : The volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

Explanation :

The balanced chemical reaction will be:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

First we have to calculate the moles of Al.

$\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}$

Molar mass of Al = 27 g/mole

$\text{Moles of }Al=\frac{55g}{27g/mol}=2mole$

Now we have to calculate the moles of $O_2$ .

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of $O_2$

So, 2 mole of Al react with $\frac{3}{4}\times 2=0.75\times 2$ moles of $O_2$

Now we have to calculate the volume of $O_2$ consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of $O_2$ occupies 22.4 liter volume of $O_2$ gas

So, $0.75\times 2$ mole of $O_2$ occupies $0.75\times 2\times 22.4$ liter volume of $O_2$ gas

Therefore, the volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

3 0

3 years ago

1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →

Leno4ka [110]

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

As 80 g NaOH produces 106 g of Na₂CO₃.

Therefore 5 g NaoH will produce ;

106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

brainly.com/question/11848702

#SPJ1

5 0

2 years ago

A balloon that has a volume of 9.45L of helium is released at the ground where the pressure is 760mmHg. What is the volume of th

max2010maxim [7]

Answer:

What is the volume of the balloon when it rises in the atmosphere to a place where pressure is 640mmHg

4 0

3 years ago

Find the theoretical oxygen demand for the

never [62]

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid, $[CH3COOH]$ = 200 mg/L

$CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O$

$ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH$

Based on the reaction stoichiometry:

mass of $CH3COOH$ = 60 g

mass of $O2$ = 2(32) = 64 g

$ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L$

b) Given:

Concentration of ethanol, $[C2H5OH]$ = 30 mg/L

$C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O$

$ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH$

Based on the reaction stoichiometry:

mass of $C2H5OH$ = 46 g

mass of $O2$ = 3(32) = 96 g

$ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L$

c) Given:

Concentration of sucrose, $[C12H22O11]$ = 50 mg/L

$C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O$

$ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11$

Based on the reaction stoichiometry:

mass of $C12H22O11$ = 342 g

mass of $O2$ = 12(32) = 384 g

$ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L$

7 0

3 years ago