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jenyasd209 [6]
2 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]2 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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Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 460^oC = 5.8\times 10^{-6}s^{-1}

K_2 = rate constant at T_2 = 4\times K_1

Ea = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 460^oC=273+460=733K

T_2 = final temperature = ?

Now put all the given values in this formula, we get:

\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]

T_2=791.1K

Therefore, the final temperature would be, 791.1 K

4 0
2 years ago
A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it
patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

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P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0
1 year ago
93.2 mL of a 2.03 M potassium fluoride (KF) solution
Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

  • Initial volume (V₁): 93.2 mL
  • Initial concentration (C₁): 2.03 M
  • Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

93.2mL \times \frac{1L}{1000mL} = 0.0932 L

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

V_2 = 0.0932L + 3.92 L = 4.01L

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M

3 0
2 years ago
Read 2 more answers
What is the molal concentration of a solution made by dissolving 34.2 g of sucrose, c12h22o11 (molar mass 342.34 g/mol), in 125
Anna007 [38]
Molality is the number of moles of solute in 1 kg of solvent
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7 0
2 years ago
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belka [17]

Answer : The molarity of solution is, 1.00 M

Explanation : Given,

Moles of KNO_3 = 0.500 mol

Volume of solution = 0.500 L

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Formula used :

\text{Molarity}=\frac{\text{Moles of }KNO_3}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

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