Given that E°_red = -0.40 V for Cd²⁺/Cd at 25 °C, find E° and E for the concentration cell expressed using shorthand notation below: Cd(s)∣ Cd²⁺(1.0 × 10^-5 mol L^-1)∣∣ Cd²⁺(0.100 mol L^-1)∣ Cd(s)

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The shorthand notation of the concentration cell is: Cd(s) ∣ Cd²⁺(1.0 × 10^-5 mol L^-1) ∣∣ Cd²⁺(0.100 mol L^-1) ∣ Cd(s) Using the Nernst equation, we can calculate the cell potential: E = E° - (RT/nF)ln(Q) where R = gas constant = 8.314 J/K·mol T = temperature in Kelvin = 298 K n = number of electrons transferred = 2 (since Cd²⁺ gains two electrons to become Cd) F = Faraday constant = 96,485 C/mol Q = reaction quotient For the anode half-cell (left), Q = [Cd2+]/[Cd²⁺] For the cathode half-cell (right), Q = [Cd2+]/[Cd²⁺] = (1.0 × 10^-5)/(0.100) Substituting the values into the equation: E = -0.40 V - (8.314 J/K·mol / (2 × 96,485 C/mol))ln([Cd2+]/[Cd²⁺]) = +0.0789 V Therefore, the answer is B) E° = 0.00 V and E = +0.0789 V.

Given That E°red = -0

Given That E°_red = -0