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jenyasd209 [6]
3 years ago
10

A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed

of copper in a 1.0 M copper(II) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C. Standard reduction potentials can be found here.
Chemistry
1 answer:
s2008m [1.1K]3 years ago
6 0

<u>Answer:</u> The standard potential for the given cell is 0.89 V

<u>Explanation:</u>

The standard reduction potentials for zinc and copper are:

E^o_{(Cu^{2+}/Cu)}=+0.13V\\E^o_{(Zn^{2+}/Zn)}=-0.76V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, copper will undergo reduction reaction will get reduced.

Zinc will undergo oxidation reaction and will get oxidized.

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

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

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.13-(-0.76)=0.89V

Hence, the standard potential for the given cell is 0.89 V

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1. Calculate the molarity when 403 g MgSO4 is dissolved to make a 5.25 L solution. Round to two significant digits.
lisov135 [29]

Answer:

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

4. Molarity of NaCl= 2.3 M

5. Mass of C₁₂H₂₂O₁₁ = 4.1 g

6. Mass of C₁₀H₈ (naphthalene) = 77 g

7. mass of ethanol = 11 g

8. Mass of DDT = 0.011 mg

Explanation:

Ans 1.

Data given

mass of MgSO₄ = 403

Volume of solution = 5.25 L

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of MgSO₄ = 24 + 32 + 4(16)

Molar mass of MgSO₄ = 120 g/mol

Put values in equation 1

             no. of moles = 403 g / 120 g/mol

             no. of moles = 3.36 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 3.36 mol / 5.25 L

           M = 0.64

So, round the figure to two significant digits.

Molarity of MgSO₄ = 0.64 M

_____________________

Ans 2.

Data given

mass of AgNO₃ = 1.24 g

Volume of solution = 125 mL

convert mL to L

1000 mL = 1 L

125 mL = 125/1000 = 0.125

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of AgNO₃ = 108 + 14 + 3(16)

Molar mass of AgNO₃ = 170 g/mol

Put values in equation 1

             no. of moles = 1.24 g / 170 g/mol

             no. of moles = 0.0073 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 0.0073 mol / 0.125 L

           M = 0.06

So, round the figure to two significant digits.

Molarity of AgNO₃ = 0.06 M

______________________

Ans 3.

Data Given:

volume of acetic acid = 500 mL

% solution of acetic acid (M)= 50%

volume of acetic acid needed = ?

Solution:

formula used

    percent of solution = volume of solute/ volume of solution x 100

Put Values in above formula

                        50 % = volume of solute / 500 mL x 100

Rearrange the above equation

                   volume of solute =   50 x 500 mL /100

                   volume of solute =   250 mL

So, volume of acetic acid = 250 mL

______________________

Ans 4

Data Given:

Molarity of NaCl (M1) = 6 M

Volume of NaCl (V1) = 750 mL

convert mL to L

1000 ml = 1 L

750 ml = 750/1000 = 0.75 L

Volume of dilute solution (V2)= 2 L

Molarity of dilute solution (M2) = ?

Solution:

Dilution Formula will be used

                M1V1 = M2V2 . . . . . . (1)

Put values in equation 1

               6  M x 0.75 L = M2 x 2 L

Rearrange the above equation

               M2 = 6 M x 0.75 L / 2 L

               M2 = 2.25 M

So, round the figure to two significant digits.

Molarity of NaCl= 2.3 M

_________________________

Ans 5.

Data Given:

Molarity of C₁₂H₂₂O₁₁ = 0.16 M

Mass of C₁₂H₂₂O₁₁ (m) = ?

Volume of dilute solution = 75 mL

convert mL to L

1000 ml = 1 L

75 ml = 75/1000 = 0.075 L

Solution:

As we know

            Molarity = no.of moles/ liter of solution . . . . . . .(1)

we also know that

           no.of moles = mass in grams / molar mass . . . . . (2)

Combine both equation 1 and 2

            Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

Put values in equation in equation 3

              0.16 M = (mass in grams / 342 g/mol) / 0.075 L

Rearrange the above equation

         mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol

         mass in grams = 4.104 g

So, round the figure to two significant digits.

Mass of C₁₂H₂₂O₁₁ = 4.1 g

____________________

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