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3 years ago

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The balanced half-reaction in which sulfate ion is reduced to sulfite ion is a ________ process. Group of answer choices a) four

-electron b) six-electron c) two-electron d) one-electron e) three-electron.

1 answer:

Aleks [24]3 years ago

7 0

Answer:

The balanced half-reaction in which sulfate ion is reduced to sulfite ion is a two electron process

Explanation:

In the sulfate anion, S, acts with +6 in the oxidation number

SO₄⁻²

This is the anion from the sulfuric acid, when it lost the 2 protons

H₂SO₄ → 2H⁺ + SO₄⁻²

In the sulfite anion, S, acts with +4 in the oxidation number.

SO₃⁻²

This is the anion, generated when the sulfurose acid lost the 2 protons.

H₂SO₃ → SO₃⁻² + 2H⁺

In acidic medium

2H⁺ + SO₄⁻² + 2e⁻ → SO₃⁻² + H₂O

In basic medium

H₂O + SO₄⁻² + 2e⁻ → SO₃⁻² + 2OH⁻

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An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

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3 years ago

What volume of 0.200 M H2SO* will be needed to neutralize<br><br> 50.00 mL of 0.178 MNaOH?

mixas84 [53]

Answer:

10.00 mL

Explanation:

Please, do not forget to write the dot and 2 of Zeros after the number 10 in the answer!

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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What is mandrake tuber

Rudiy27

It's botanical name would be Mandragora officianarum, and it has a humanoid shape. Its a murderous plant that grows from blood that grows in Mediterranean.

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What is the difference between a plain and a plateau? A. Only a plain is flat, a plateau is steep mountainside. B. Both are flat

Alenkinab [10]

Answer:

B

Explanation:

<em>Both plain and plateau have flat surfaces. However, a plain is located in a low-lying area while a plateau is located on an elevated area. In essence, a plateau can be viewed as an elevated plain or a plain that is bordered by cliffs.</em>

The correct option is B.

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2 years ago

What will be the pressure of 2.00 mol of an ideal gas at a temperature of 20.5 degrees Celsius and a volume of 62.3L?

erastova [34]

.774atm

First, look at what you have and look at the equations you can use to solve this problem. The best equation would be PV=nRT.

P being pressure, V being volume, n being moles, R being the gas constant, and T being temperature.

Before you start doing any of the math, make sure of two things. Since you're looking for pressure, you'll need a gas constant. When I did the problem, I used the gas constant of atm or atmospheres which is .0821.

Also! Remember to always convert celsius into kelvin, to do this, add 273 to the given celsius degree. After this is all set and done, your equation should look like this:

P = $\frac{2 x .0821 x 293.5}{62.3}$

The reason that the equation is divided by the volume is due to the fact that you need to isolate the variable or pressure.

Multiply everything on the top and divide by the bottom and you should receive the final answer of .774atm.

Hope this helps!

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3 years ago