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Karolina [17]
4 years ago
10

What is the shorthand notation (i.e., line notation) that best represents a complete electrochemical cell for the following galv

anic cell reaction?
Sn^2+ (aq) + F2 (g) → Sn^3+ (aq) + 2F^− (aq)
O Pt(s) | Sn^2+ (aq), Sn^3+ (aq) || F2 (g) F^-(aq)| C(s)
O F2(g)|F^- (aq) || Sn^2+(aq) | Sn^3+ (aq)
O Sn(s) | Sn^2+(aq) || Sn^4+ (aq) F2 (g)|F^- (aq)| C(s)
O Sn^2+ (aq) | Sn^3+ (aq) || F2 (g)|F^- (aq)
O Pt(s) | Sn^4+ (aq), Sn2^+ (aq), F2 (g) || F^- (aq) | C(s)
Chemistry
1 answer:
ruslelena [56]4 years ago
5 0

Answer:

Sn^2+(aq)/Sn^3+(aq)//F2(g)/2F-(aq)

Explanation:

In writing the shorthand notation for an electrochemical cell, the oxidation half cell is shown on the left hand side and the reduction half cell is shown on the right hand side. The oxidation half equation reflects electron loss while the reduction half equation reflects electron gain.

You might be interested in
The ph of a solution containing 0.818 M acetic acid<br> (ka1.76*10-5) ans 0.172 M sodium acetate is?
kvasek [131]

Explanation:

It is known that the relation between pH and pK_{a} is as follows.

              pH = pK_{a} + log \frac{[salt]}{[acid]}

and,     pK_{a} = -log K_{a}

Hence, first we will calculate the value of pK_{a} as follows.

                   pK_{a} = -log K_{a}

                               = -log (1.76 \times 10^{-5}

                               = 4.75

Now, we will calculate the value of pH as follows.

              pH = pK_{a} + log \frac{[\text{sodium acetate}]}{\text{acetic acid}}

                    = 4.75 + log \frac{0.172}{0.818}      

                    = 4.75 + (-0.677)

                    = 4.07

Therefore, we can conclude that the pH of given solution is 4.07.

5 0
4 years ago
Consider the following reversible reaction.
andriy [413]

Answer:

No one is correct. The correct expression is:

Keq = [H₂]²  . [O₂]² / [H₂O]²

Explanation:

To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.

The balance reaction is:

<u>2</u> H₂O (g)  ⇄  <u>2</u> H₂ (g)  +  O₂ (g)

Keq = [H₂]²  . [O₂]  / [H₂O]²

In opposite side: <u>2</u> H₂ (g)  +  O₂ (g)   ⇄  <u>2</u> H₂O (g)

Keq =  [H₂O]² / [H₂]²  . [O₂]  

6 0
4 years ago
1. An oxide of chromium is found to have the following % composition: 68.4% Cr
abruzzese [7]

Answer:

Empirical formula is Cr₂O₃.

Explanation:

Given data:

Percentage of Cr = 68.4%

Percentage of O = 31.6%

Empirical formula = ?

Solution:

Number of gram atoms of Cr = 68.4 / 52 = 1.3 2

Number of gram atoms of O = 31.6 / 16 = 1.98

Atomic ratio:

                            Cr               :         O

                           1.32/1.32     :       1.98/1.32

                               1              :        1.5

Cr : O = 1 :  1.5

Cr : O = 2(1 : 1.5)

Empirical formula is Cr₂O₃.

6 0
3 years ago
Magnetism is believed to be caused by the alignment of small, numerous sub-units called
77julia77 [94]

Answer:

Magnetism is believed to be caused by the alignment of small, numerous sub-units called : <em><u>Domains</u></em>

<em><u></u></em>

Explanation:

Domains : A magnetic domain is the region in which in which magnetic field of the atoms are grouped together and aligned.

  • In unmagnetized material all the magnetic Domains  point in different direction.
  • In magnetised material (ferromagnets , antiferromagnets) , The Domains point in a particular( fixed Pattern) direction.

5 0
3 years ago
1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%
Tatiana [17]

Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

3 0
3 years ago
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