= Solution
On the stated $1\,\mathrm{m^2}$ area, the total <electric current> is $I=200\,\mathrm A$. The known <ohmic polarization> is $IR_\Omega=1.20\,\mathrm V$. Applying the supplied anodic <Tafel equation> gives
$$
\eta_a=0.295\log_{10}(200/1)=0.6788\,\mathrm V.
$$
Assume 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> magnitude
$$
|\eta_{c,\rm conc}|=\frac{RT}{2F}\ln41=0.04765\,\mathrm V.
$$
Relative to the local surface equilibrium potential, the <cathode> also needs an <activation overpotential>. Neglecting anodic <concentration> polarization, the separated-loss accounting is
$$
\boxed{V_{\rm app}\simeq1.9265\,\mathrm V+|\eta_{c,\rm act}|}.
$$
To calculate that activation term, adopt a single effective two-electron <Butler-Volmer equation> with constant transfer coefficients satisfying $\alpha_a+\alpha_c=1$, no anodic <concentration> polarization and <reduction> proportional to the surface $\mathrm{Cu}^{2+}$ <concentration>. The supplied anodic <Tafel slope> then determines the reverse slope through
$$
\alpha_a=\frac{RT\ln10}{2Fb}\simeq0.10016,\qquad\alpha_c=1-\alpha_a\simeq0.89984.
$$
With $f=2F/(RT)$ and cathodic <overpotential> $\eta_c$ referenced to bulk equilibrium,
$$
\frac{j_c}{j_0}=\exp(\alpha_a f\eta_c)-\frac{c_s}{c_b}\exp(-\alpha_c f\eta_c).
$$
Solving at $j_c=-200\,\mathrm{A\,m^{-2}}$ gives $\eta_c\simeq-0.12854\,\mathrm V$, including both <concentration> and activation effects. Thus
$$
\boxed{V_{\rm app}\simeq1.20+0.67880+0.12854=2.0073\,\mathrm V\simeq2.01\,\mathrm V.}
$$
The $0.04765$ V <concentration> loss must not be added again. The kinetic assumption matters: the supplied slope is for <copper> <oxidation>, and an <exchange current density> alone does not specify every cathodic current-potential relation. A multistep <copper> reaction need not obey the effective single-step relation used for this estimate.
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