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A kinetic model for the net electrode current near and away from equilibrium. For a single effective reaction with consistent activities and ${\alpha }_a+{\alpha }_c=1$, $j=j_0[\exp ({\alpha }_{a}nF\eta /RT)-\exp (-{\alpha }_{c}nF\eta /RT)]$. Near equilibrium it gives $j\simeq j_{0}nF\eta /(RT)$; at sufficiently large overpotential one exponential gives the Tafel equation. Multistep reactions can require a different kinetic interpretation.

On the stated $1{\mathrm{m}}^2$ area, the total electric current is $I=200\mathrm{A}$. The known ohmic polarization is $I{R}_{\Omega }=1.20\mathrm{V}$. Applying the supplied anodic Tafel equation givesAssume the quoted limiting current uses the same cathodic area convention. A steady diffusion-film model gives $c_s/c_b=1-j/j_L=1/41$. The Nernst equation then makes the cathodic concentration overpotential magnitudeRelative to the local surface equilibrium potential, the cathode also needs an activation overpotential. Neglecting anodic concentration polarization, the usual separated-loss accounting isThe printed $b$ is explicitly for copper oxidation, not a separately specified reduction slope. An exchange current density by itself does not determine the whole cathodic current-potential relation. TODO: a unique applied voltage requires a cathodic kinetic law, or an explicit additional relation between anodic and cathodic kinetics.

For a concrete conditional completion, assume one effective two-electron Butler-Volmer equation with ${\alpha }_a+{\alpha }_c=1$, no anodic concentration polarization and reduction proportional to the surface ${\mathrm{Cu}}^{2+}$ concentration. The anodic slope implies ${\alpha }_a=2.303RT/(2Fb)=0.10016$ and ${\alpha }_c=0.89984$. With $f=2F/(RT)$ and cathodic overpotential ${\eta }_c$ referenced to bulk equilibrium,Solving at $j_c=-200\mathrm{A}{\mathrm{m}}^{-2}$ gives ${\eta }_c=-0.12854\mathrm{V}$, including both concentration and activation effects. Therefore this explicitly assumed model gives $V_{\mathrm{app}}=1.20+0.67880+0.12854=2.0073$ V. The $0.04765$ V concentration loss must not be added again. A multistep copper reaction need not obey this effective single-step relation, so 2.0073 V is not uniquely entailed by the printed data.

First hold the production rate fixed, so total electric current remains $200\mathrm{A}$. If only the impure anode receives a factor $10^5$ increase in accessible area, its local current density becomes $0.002\mathrm{A}{\mathrm{m}}^{-2}$, far below $j_0=1\mathrm{A}{\mathrm{m}}^{-2}$. Extrapolating the large-current Tafel equation would give a negative anodic overpotential, which is physically the wrong regime. Near equilibrium the Butler-Volmer equation instead gives, in the same conditional two-electron model,The anode activation loss nearly disappears; the 1.20 V ohmic loss does not disappear. The cathode still operates close to its original transport limit if it is unchanged. The conditional kinetic model gives $V_{\mathrm{app}}\simeq 1.3286$ V for anode-only enlargement. Its anodic potential returns close to 0.34 V, making retention of noble silver and gold more favorable, while less noble nickel dissolution and lead sulfate formation remain relevant.

If both active areas increase by $10^5$ and their independent mass-transfer capacity scales by the same factor, both local kinetic losses become small and the conditional prediction is about $1.20005$ V. Porous internal area does not generally multiply an external diffusion-limited flux by $10^5$: pore access, ion depletion, tortuosity, electrical connectivity and slime blockage must be checked. If the specified 200 A/m² is instead held per true enlarged area, total current rises to $2\times 10^7$ A and the unchanged electrolyte resistance would give $1.2\times 10^5$ V, so this is an entirely different operating condition.

TODO: an unconditional revised voltage requires the missing cathodic law and a specified area/current/mass-transfer convention. The robust conclusion at fixed total current is reduced local activation polarization on the genuinely accessible enlarged surface, with an unchanged ohmic floor unless cell geometry or conductivity also changes. Increased geometric area alone does not prove a proportionate decrease in energy cost or improved purity.

The reversible voltage follows from the two-electron hydrogen reaction:At the specified geometric current density, the magnitude of each activation loss follows its Tafel equation, $\eta =b{\log }_{10}(j/j_0)$:Both losses reduce the delivered fuel cell voltage. Therefore $V=0.36048\mathrm{V}$ and $P/A=jV=3.60\mathrm{W}{\mathrm{m}}^{-2}$. The very small oxygen exchange current is responsible for most of the activation loss. The calculation uses current densities on a common area basis, unit water activity and the stated gas conditions, and neglects the losses the problem excludes.

At unchanged intrinsic catalyst kinetics, increasing the electrochemically accessible true area by $100$ multiplies exchange current per geometric area by $100$. Equivalently, the local current density at a fixed total geometric current falls by $100$. Each Tafel equation activation loss decreases by $b{\log }_{10}100=2b$:Thus $V_{\mathrm{new}}=0.59948\mathrm{V}$ and $P_{\mathrm{new}}/A=5.99\mathrm{W}{\mathrm{m}}^{-2}$, an increase of $2.39\mathrm{W}{\mathrm{m}}^{-2}$, or about $66.3\%$. Power does not increase a hundredfold, because the voltage improvement is logarithmic and the specified geometric current is unchanged. This assumes both electrodes gain accessible active area, the original exchange currents use the original area basis, and the porous structure does not introduce additional transport or Ohmic losses.