Half-cell / reduction potential?

• eraemia
In summary, the conversation discusses the potential difference between half-cell Xn+/X and half-cell Yn+/Y, as well as the potential difference between Yn+/Y and Cu2+. It also mentions the reduction potential of Cu2+ + 2 e- right arrow Cu and asks for the reduction potential of Xn+ + n e- right arrow X relative to the standard hydrogen half-cell. The proposed solution is 0.25 V.
eraemia
Could anyone give me some pointers regarding this problem?

Half-cell Xn+/X is connected with half-cell Yn+/Y and the potential difference is found to be 0.12V with Y being the positive electrode. Yn+/Y is connected with Cu2+ and the potential difference is found to be 0.13V with Cu being the positive electrode. If Cu2+ + 2 e- right arrow Cu is assigned the potential 0.34V versus the standard hydrogen electrode, what is the reduction potential of Xn+ + n e- right arrow X relative to the standard hydrogen half-cell?

My tentative solution is 0.12 V + 0.13 V = 0.25 V. Does that look right?

Thanks!

Whats all this Xn+/X malarchy? Is that the representation of the half cell? If so please make it slightly more clear what's being read my understanding is:

X|X+||Y+|Y

but I am not really getting it

I can provide some guidance on this problem. The reduction potential of a half-cell is a measure of its tendency to gain electrons and undergo reduction. It is typically measured in volts (V) and can be used to compare the reactivity of different half-cells.

In this problem, we are given the potential differences for two half-cells, Yn+/Y and Cu2+/Cu, relative to the standard hydrogen electrode (SHE). The SHE is often used as a reference point for measuring reduction potentials and is assigned a value of 0.00V. The reduction potential of Cu2+ + 2 e- → Cu is given as 0.34V, meaning that this half-cell has a stronger tendency to undergo reduction compared to the SHE.

To determine the reduction potential of Xn+ + n e- → X, we can use the Nernst equation: E = E° - (RT/nF)ln(Q). In this equation, E is the measured potential difference, E° is the standard reduction potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient.

Using the potential difference of 0.12V for the Xn+/X half-cell and the potential difference of 0.13V for the Yn+/Y half-cell, we can calculate the reaction quotient Q as Q = (0.13V)/(0.12V) = 1.08.

Now, substituting the given values into the Nernst equation and solving for E°, we get: E° = E + (RT/nF)ln(Q) = 0.25V + (0.0257 V/K)(ln(1.08)) = 0.25V + 0.002V = 0.252V.

Therefore, the reduction potential of Xn+ + n e- → X relative to the standard hydrogen half-cell is 0.252V. Your solution of 0.25V is very close and can be considered correct. I hope this helps!

1. What is a half-cell?

A half-cell is a component of an electrochemical cell that consists of a single electrode and a solution containing the ions of the electrode's corresponding element. It is used to measure the reduction potential of a specific element.

2. What is reduction potential?

Reduction potential, also known as oxidation-reduction potential, is a measure of the tendency of a chemical species to undergo reduction. It is measured in volts and indicates the strength of the oxidizing or reducing agent in a chemical reaction.

3. How is the reduction potential of a half-cell determined?

The reduction potential of a half-cell is determined by measuring the voltage between the half-cell and a standard hydrogen electrode. This voltage is known as the standard reduction potential and is used as a reference point for all other reduction potentials.

4. What is the Nernst equation and how is it used in relation to half-cell potential?

The Nernst equation is a mathematical formula that relates the reduction potential of a half-cell to the concentrations of the species involved in the reaction. It is used to calculate the potential of a half-cell under non-standard conditions, such as different concentrations or temperatures.

5. How does temperature affect half-cell potential?

Temperature has a direct effect on the reduction potential of a half-cell. As temperature increases, the reduction potential also increases, and vice versa. This is due to the fact that temperature affects the rate of chemical reactions, and reduction potentials are a measure of the rate of reduction reactions.

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