Electrochemistry - Voltaic Cell

In summary: Therefore, the correct answer is a) 3.142 V; b) 2.958 V; c) zero. In summary, the value of E for the voltaic electrochemical cell at 25°C is 3.142V, at the stage when 1.2g of silver has been deposited, it is 2.958V, and at equilibrium, it is zero.
  • #1
iharuyuki
16
1
Please post this type of questions in HW section using the template.
A voltaic electrochemical cell consists of a silver electrode in contact with 200 mL of 0.1 M AgNO3 and a magnesium electrode in contact with 300 mL of a 0.1 M Mg(NO3)2 solution. We have the following data: Ag+ + e- --> Ag (E° = 0.8 V) Mg2+ + 2e- --> Mg (E° = -2.372 v)

a) Calculate the value of E for the cell at 25°C
b) A current is drawn from the cell until 1.2 g of silver has been deposited at the silver electrode. Calculate the value of E for the cell at this stage.
c) Determine E for the cell once the cell has reached equilibrium.

a a) 3.142 V; b) 2.958 V; c) zero
b a) 3.172 V; b) 3.122 V; c) 3.122 V
c a) 3.172 V; b) 3.122 V; c) 3.172 V
d a) 3.142 V; b) 2.958 V; c) 2.958 V

Attempt at solution:
Overall reaction:
2Ag+ + Mg(s) -> 2 Ag(s) + Mg2+ E0cell = 3.172

Q = [Mg2+]/[Ag+]^2 solid phase components not included

For part a) Nerst equation with E0cell = 3.172 V and finding Q with the given concentrations above yields E25C = 3.142V So b and c are eliminatedI'm not quite sure how to proceed from there. Is it possible for Ecell at equilibrium to be 0? If not, A is eliminated as well.

Thank you for the help, and sorry if I've broken any etiquette. It's my first time posting here.
 
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  • #2
Yes, it is possible for the cell potential at equilibrium to be zero. This occurs when the reaction has reached its equilibrium point and no further net reaction can occur. To solve for part b), we need to calculate the number of moles of silver that has been deposited. Using the molar mass of silver (107.87 g/mol), we can calculate the number of moles of silver deposited as 1.2g/107.87g/mol = 0.011 mol. We then need to calculate the new concentrations of Ag+ and Mg2+ in the cell. The new concentration of Ag+ will be 0.1M - 0.011mol/0.3L = 0.0989M and the new concentration of Mg2+ will be 0.1M + 0.011mol/0.3L = 0.1009M. Using these new concentrations in the Nernst equation with E0cell = 3.172V, we can calculate E = 3.122V. For part c), since the reaction has reached its equilibrium point, the cell potential is zero.
 

Related to Electrochemistry - Voltaic Cell

1. What is a Voltaic Cell?

A Voltaic Cell is an electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction. It consists of two half-cells, an anode and a cathode, connected by a conductive pathway and a salt bridge.

2. How does a Voltaic Cell work?

A Voltaic Cell works through a redox reaction, where one half-cell undergoes oxidation (loses electrons) and the other half-cell undergoes reduction (gains electrons). This creates a flow of electrons from the anode to the cathode, producing an electrical current.

3. What is the difference between a Voltaic Cell and an electrolytic cell?

The main difference between a Voltaic Cell and an electrolytic cell is the source of energy. In a Voltaic Cell, the energy comes from the spontaneous redox reaction, while in an electrolytic cell, an external source of energy is required to drive the non-spontaneous redox reaction.

4. What factors affect the voltage of a Voltaic Cell?

The voltage of a Voltaic Cell is affected by the types of metals used as electrodes, their concentrations, and the temperature. The use of different electrolytes and the surface area of the electrodes also play a role in determining the voltage.

5. What are some practical applications of Voltaic Cells?

Voltaic Cells have many practical applications, including powering electronic devices, such as calculators and watches, and providing backup power for emergency lighting. They are also used in batteries for vehicles and renewable energy sources, such as solar panels and wind turbines.

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