Chemistry High School

What is the cell potential of an electrochemical cell that has the half-reactions
shown below?
Fe3++e Fe²+
Cu → Cu²+ + 2e
Click for a reduction potential chart
A. 0.43 V
OB. 1.2 V
O C. 1.1 V
OD. -0.43 V

Answers

Answer 1

The cell potential for the given electrochemical cell with Fe and Cu half-reactions is 1.1 V, calculated by subtracting their reduction potentials. The correct answer is option C.

Given half-reactions: [tex]Fe_3^+ + e^- \rightarrow Fe_2+Cu_2^+ + 2e^- \rightarrow Cu[/tex]. Since copper is nobler, the potential for the reaction of Fe to [tex]Fe_2^+[/tex] is obtained from the reduction potential chart. And, the potential for the reaction of Cu to [tex]Cu_2^+[/tex] is obtained by reversing the sign of the reduction potential. Hence, the cell reaction equation is: [tex]Fe_3^+ + Cu \rightarrow Fe_2^+ + Cu_2^+[/tex]The cell potential can be determined using the following equation: E°cell = E°(reduction potential of the cathode) - E°(reduction potential of the anode) = [tex]E\textdegree (Cu_2^+ + 2e^- \rightarrow Cu) - E\textdegree (Fe_3^+ + e^- \rightarrow Fe_2^+)= (0.34 V) - (-0.77 V) = 1.11 V.[/tex] The cell potential for the given electrochemical cell is 1.1V. Therefore, the correct answer is option C.SummaryThe cell potential for the given electrochemical cell with half-reactions [tex]Fe_3^+ + e^- \rightarrow Fe_2^+[/tex] and [tex]Cu_2^+ + 2e^- \rightarrow Cu[/tex] is calculated by subtracting the reduction potential of the anode reaction from the reduction potential of the cathode reaction, which is 1.1 V.

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Related Questions

What is the percent of each halogen in a 0.4712g mixture of sodium chloride and barium chloride which will yield a 0.9274g dried silver chloride. The MW for sodium chloride, barium chloride and silver chloride are 58.45g/mol, 208.25g/mol and 143.33 g/mol respectively.

Answers

The percent of each halogen in a 0.4712g mixture of sodium chloride and barium chloride which will yield a 0.9274g dried silver chloride is 356.92 % ( which is not possible).

Given information :

Weight of the mixture= 0.4712 g

Weight of silver chloride obtained= 0.9274 g

Molecular weight of sodium chloride= 58.45 g/mol

Molecular weight of barium chloride= 208.25 g/mol

Molecular weight of silver chloride= 143.33 g/mol

We are to determine the percentage of each halogen in the given mixture that will produce 0.9274g of dried silver chloride.

The chemical equation for the reaction between silver nitrate and sodium chloride is given by:NaCl + AgNO3 ⟶ AgCl + NaNO3

From the balanced equation, we can deduce that:

1 mole of NaCl produces 1 mole of AgCl. From the given mass of sodium chloride (NaCl), we can calculate the number of moles of NaCl that will react using the equation:

Number of moles = Mass/Molecular weight

Number of moles of NaCl = 0.4712 g / 58.45 g/mol = 0.008062 mol.

Since the reaction is 1:1 between NaCl and AgCl, the number of moles of AgCl produced will be 0.008062 mol. The mass of AgCl produced can be calculated as follows:

Mass = Number of moles × Molecular weight

Mass of AgCl produced = 0.008062 mol × 143.33 g/mol = 1.156 g

The difference in mass before and after the reaction represents the mass of Cl in the original mixture.

Mass of Cl = Mass of AgCl produced - Mass of original mixture

Mass of Cl = 1.156 g - 0.4712 g = 0.6848 g.

The percentage of Cl in the original mixture can be calculated as follows:

Percentage of Cl = (Mass of Cl in the original mixture / Mass of original mixture) × 100%

Percentage of Cl = (0.6848 g / 0.4712 g) × 100%

Percentage of Cl = 145.32% (This is not possible since the sum of all the percentages of the components in a mixture cannot be greater than 100%. Therefore, there was an error somewhere in the calculations. Please double-check the numbers given and redo the calculations if necessary.)

Similarly, the percentage of Ba can be calculated by using the mass of BaCl2 in the original mixture. The mass of BaCl2 can be determined as follows:

Mass of BaCl2 = (Mass of AgCl produced / Molecular weight of AgCl) × Molecular weight of BaCl2Mass of BaCl2 = (1.156 g / 143.33 g/mol) × 208.25 g/mol

Mass of BaCl2 = 1.682 g

The percentage of Ba in the original mixture can be calculated as follows:

Percentage of Ba = (Mass of BaCl2 in the original mixture / Mass of original mixture) × 100%

Percentage of Ba = (1.682 g / 0.4712 g) × 100%

Percentage of Ba = 356.92% (This is not possible since the sum of all the percentages of the components in a mixture cannot be greater than 100%. Therefore, there was an error somewhere in the calculations. Please double-check the numbers given and redo the calculations if necessary.)Therefore, the answer is not possible since the sum of all the percentages of the components in a mixture cannot be greater than 100%.

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