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Nernst Equation for Lithium ion battery
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Kasturi Khatun
Nernst Equation for Lithium ion battery
The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.
The same reasoning may be done on the Nernst equation at the cathode : $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.
The overall voltage of the battery $\Delta E$ can be written : $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.
The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.
The same reasoning may be done on the Nernst equation at the cathode : $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.
The overall voltage of the battery $\Delta E$ can be written : $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.
After discharge and charge the voltage jumps or drops immediately due to the activation and ohmic overpotentials. But the change due to the mass transport loss/concentration loss takes time. This change is due to the concentration diffusion but the Li+ ions concentration at the electrodes are not in the overall voltage. Are you saying the solid concentration is changing with time while no current is flowing?More
The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.
The same reasoning may be done on the Nernst equation at the cathode : $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.
The overall voltage of the battery $\Delta E$ can be written : $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.
The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.
The same reasoning may be done on the Nernst equation at the cathode : $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.
The overall voltage of the battery $\Delta E$ can be written : $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.
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