Calculate the potential difference (in volts) between two half-cells when one half-cell contains a zinc (Zn) electrode immersed in a 1.0 M Zn²⁺ solution, and the other half-cell contains a silver (Ag) electrode immersed in a 0.10 M Ag⁺ solution. (Use standard reduction potentials)
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To calculate the potential difference between two half-cells using standard reduction potentials, we need to use the Nernst equation:
E = E° - (0.0592 V / n) * log(Q)
Where:
- E is the potential difference between the two half-cells.
- E° is the standard reduction potential of the half-reaction.
- n is the number of electrons transferred in the half-reaction.
- Q is the reaction quotient, calculated by dividing the product of the concentrations of the oxidized and reduced species by the stoichiometric coefficients.
First, let's determine the standard reduction potentials for the given half-reactions:
- For the Zn half-reaction: Zn²⁺(aq) + 2e⁻ -> Zn(s)
- E° = -0.76 V
- n = 2
- For the Ag half-reaction: Ag⁺(aq) + e⁻ -> Ag(s)
- E° = 0.80 V
- n = 1
Now, let's calculate the reaction quotient, Q:
Q = [Zn²⁺] / [Ag⁺]
Q = (1.0 M) / (0.10 M)
Q = 10
We can now substitute the values into the Nernst equation for each half-reaction:
E(Zn) = -0.76 V - (0.0592 V / 2) * log(10)
E(Zn) = -0.76 V + (0.0296 V) * log(10)
E(Zn) ≈ -0.76 V + 0.0296 V
E(Zn) ≈ -0.73 V
E(Ag) = 0.80 V - (0.0592 V / 1) * log(10)
E(Ag) = 0.80 V - (0.0592 V) * log(10)
E(Ag) ≈ 0.80 V - 0.0296 V
E(Ag) ≈ 0.77 V
Finally, we can calculate the potential difference:
E = E(Zn) - E(Ag)
E = -0.73 V - 0.77 V
E ≈ -1.50 V
Therefore, the potential difference between the two half-cells is approximately -1.50 volts.
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