For the electrochemical cell below, the half-cell on the left is the anode and the cell potential is 0.344 V at 25 °C. What is the concentration of H+, x, in mol L−1 ? Note: The symbol M is an abbreviation for mol L−1. Pt(s) | Cr2+(aq, 0.520 M) , Cr3+(aq, 0.700 M) || H+(aq, x M ) | H2(g, 1.10 bar) | Pt(s) Half reaction E ° / V 2 H+(aq) + 2 e− ⇌⇌ H2(g) 0.000 Cr3+(aq) + e− ⇌⇌ Cr2+(aq) −0.424 Enter a value accurate to 3 significant figures.
Question
For the electrochemical cell below, the half-cell on the left is the anode and the cell potential is 0.344 V at 25 °C. What is the concentration of H+, x, in mol L−1 ? Note: The symbol M is an abbreviation for mol L−1. Pt(s) | Cr2+(aq, 0.520 M) , Cr3+(aq, 0.700 M) || H+(aq, x M ) | H2(g, 1.10 bar) | Pt(s) Half reaction E ° / V 2 H+(aq) + 2 e− ⇌⇌ H2(g) 0.000 Cr3+(aq) + e− ⇌⇌ Cr2+(aq) −0.424 Enter a value accurate to 3 significant figures.
Solution
To solve this problem, we need to use the Nernst equation, which relates the cell potential to the concentrations of the species involved in the reaction. The Nernst equation is:
E = E° - (RT/nF) * ln(Q)
where: E is the cell potential, E° is the standard cell potential, R is the gas constant (8.314 J/(mol*K)), T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.
First, we need to write the overall cell reaction. The half-reactions are:
Cr3+(aq) + e- ⇌ Cr2+(aq) E° = -0.424 V 2 H+(aq) + 2 e- ⇌ H2(g) E° = 0.000 V
The overall cell reaction is:
Cr3+(aq) + 2 H2(g) ⇌ Cr2+(aq) + 2 H+(aq)
The standard cell potential E° for the overall reaction is the difference of the standard potentials of the cathode and anode:
E° = E°(cathode) - E°(anode) = 0.000 V - (-0.424 V) = 0.424 V
The cell potential E is given as 0.344 V. The number of electrons transferred in the reaction n is 1. The temperature T is 25 °C, which is 298.15 K in Kelvin.
We can now plug these values into the Nernst equation and solve for the concentration of H+, x:
0.344 V = 0.424 V - (8.314 J/(mol*K) * 298.15 K / (1 * 96485 C/mol)) * ln([Cr2+][H+]^2 / [Cr3+][H2])
We know that [Cr2+] = 0.520 M, [Cr3+] = 0.700 M, and [H2] = 1.10 bar. We can assume that 1 bar is approximately equal to 1 M, so [H2] = 1.10 M. We can now solve for [H+], x:
0.344 V = 0.424 V - (8.314 J/(mol*K) * 298.15 K / (1 * 96485 C/mol)) * ln((0.520 M * x^2) / (0.700 M * 1.10 M))
Solving this equation for x will give the concentration of H+ in mol L-1.
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