Calculating Ecell and Voltage Across a Galvanic Cell | Nernst Equation Problem

In summary, the conversation discusses the calculations for the open circuit potential and voltage across a Galvanic cell made of two half cells. The open circuit potential is calculated using the Nernst equation and standard electrode potentials, while the voltage across the cell is calculated using Ohm's Law. However, the calculation for the voltage in part b) is incorrect as it does not take into account the results of part a).
  • #1
madman01
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0

Homework Statement



Consider a Galvanic cell made of two half cells:
Ag+ + e− → Ag(s) +0.799 V
Cd2+ + 2e− → Cd(s) − 0.403 V

(a) Calculate the open circuit potential, when the concentration of Ag+ 0.08 mol and the concentration of the Cd2+ is 0.8 mol.

(b) Calculate the voltage across the cell, if the internal impedance of the cell is 10 Ω and you connect it to am external resistor, which is 80 Ω

Homework Equations



E = E ° - (0.0591V/n * log(K)

The Attempt at a Solution



part a

For Ag+2 (n= 1 as 1 electron participate) and E ° = +0.799 V

E = E ° - (0.0591V/n * log(1/[Ag+2])
E = +0.799V - (0.0591V/1 * log(1/[0.08])
E = +0.799V - (0.0591V * log12.5)
E = +0.799V - (0.065V)
E = +0.734VFor Cd+2 (n= 2 as 2 electron participate) and E ° = -0.403V

E = E ° - (0.0591V/n * log(1/[Cd+2])
E = -0.403V - (0.0591V/2 * log(1/[0.8])
E = -0.403V - (0.0295V * log1.25)
E = -0.403V - (0.00285)
E = -0.406V

Ecell = Oxidation potential + Reduction potential

Ecell = 0.734V + 0.406V
Ecell = 1.140Vpart b

V = I * R
1.140 = I * (10+80)
I = 1.26 * 10^-2 A

since current is same in series therefore voltage across the cell is equal to

V = I * R
V = 1.26 * 10-2 * 10
V= 1.26* 10^-1 VIs this right ?
 
Last edited:
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  • #2
No, part b) is not right.

Clue: when you calculated the voltage of the open circuit cell voltage, you used standard electrode potentials and the Nernst equation. But none of the result of part a) was incorporated in your calculations for part b) -- which is a correct calculation of *something*.
 

Related to Calculating Ecell and Voltage Across a Galvanic Cell | Nernst Equation Problem

1. What is the Nernst Equation?

The Nernst Equation is a mathematical equation that relates the standard electrode potential of a half-cell to the concentration of ions in a solution. It is used to calculate the potential of a cell at non-standard conditions, such as different concentrations or temperatures.

2. What is the significance of the Nernst Equation?

The Nernst Equation is significant because it helps us understand how changing conditions, such as concentration and temperature, affect the potential of a cell. It is also used in electrochemistry to determine the equilibrium constant of a reaction.

3. How do you use the Nernst Equation to solve a problem?

To solve a problem using the Nernst Equation, you first need to identify the half-cell reactions and their standard electrode potentials. Then, you can use the equation E = E° - (0.0592/n)log(Q) to calculate the cell potential, where E is the potential at non-standard conditions, E° is the standard potential, n is the number of electrons transferred in the reaction, and Q is the reaction quotient.

4. What are the limitations of the Nernst Equation?

The Nernst Equation assumes that the reaction is at equilibrium and that the system is ideal, which may not always be the case in real-life situations. It also does not take into account other factors that may affect the potential of a cell, such as the presence of impurities or side reactions.

5. How is the Nernst Equation related to the Nernst distribution law?

The Nernst Equation is derived from the Nernst distribution law, which states that the ratio of the concentration of a solute in two phases at equilibrium is equal to the ratio of their activities. The Nernst Equation is a specific application of this law to electrochemistry, where the concentration of ions in a solution is related to the potential of a cell.

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