Need help fast please 15 points!!!

What will be the end product for each electrode after electrolysis if the solution is concentrated aqueous sodium chloride ?

Eddi Din [679]

Answer:

Cathode: Hydrogen gas

Anode: Chlorine gas

Explanation:

① Write down the ions present in the electrolyte

Cations: Na⁺, H⁺

Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

For the discharge of cations,

- Reactivity series

(The lower the position of the cation in the reactivity series, the easier it is to be discharged)

For the discharge of anions,

- Concentration effect (look at this first)

(The more concentrated the ion, the easier for it to be discharged)

- If solution is not concentrated (dilute), look at the position of the anion on the electrochemical series.

(The lower the position of the anion on the electrochemical series, the easier of it to be discharged)

In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

In summary, here's what happened at each electrode:

Cathode

- H⁺ selectively discharged

- ionic half equation: 2H⁺ (aq) +2e⁻ → H₂ (g)

Anode

- Cl⁻ ions selectively discharged

- ionic half equation: 2Cl⁻ (aq) → Cl₂ (g) + 2e⁻

3 0

3 years ago

What is the balanced equation for H3PO4=H4P2O7+H2O

podryga [215]

Answer:

2H3PO4=H4P2O7+H2O

Explanation:

2H3PO4=H4P2O7+H2O

5 0

3 years ago

Find the volume of a box with length 25 cm, height 25 cm and width 1.0 m. Volume

mariarad [96]

The volume of a box with length 25 cm, height 25 cm and width 1.0 m is 0.0625m³.

Volume of the box which is a cuboid can be calculated by multiplying the length, breadth and height of the given box.

Volume of the box is given by the product of the length of the box, Height of the box and Breadth or width of the box.

Since, the box is a cuboid, hence the formula is given by the products of length, breadth and height.

Given,

length of the box= 25cm = 0.25m

Height of the box =25cm = 0.25m

width of the box= 1m

Volume = length × width × height of box

Volume = 0.25 × 0.25 × 1

Volume = 0.0625m³

The volume of the box is 0.0625m³.

Learn more about Volume here, brainly.com/question/23118276

#SPJ9

4 0

1 year ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

Which do you think could work to reduce air pollution and global warming

natka813 [3]

Recycling!! And also planting more trees.

Hope this helps!

5 0

3 years ago