Given that E°_red = -1.66 V for Al³⁺/ Al at 25 °C, find E° and E for the concentration cell expressed using shorthand notation below: Al(s) ∣ Al³⁺(1.0 × 10^-5 mol L^-1) ∣∣ Al³⁺(0.100 mol L^-1) ∣ Al(s)

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The shorthand notation for the concentration cell is: Al(s) | Al³⁺(1.0 × 10^-5 mol L^-1) || Al³⁺(0.100 mol L^-1) | Al(s) The anode is the half-cell with higher concentration, which is on the right side of the cell diagram: Al(s) | Al³⁺(1.0 × 10^-5 mol L^-1) || Al³⁺(0.100 mol L^-1) | Al(s) Therefore, the oxidation half-reaction is: Al(s) → Al³⁺ + 3 e- The cathode is the half-cell with lower concentration, which is on the left side of the cell diagram: Al(s) | Al³⁺(1.0 × 10^-5 mol L^-1) || Al³⁺(0.100 mol L^-1) | Al(s) Therefore, the reduction half-reaction is: Al³⁺ + 3 e- → Al(s) The standard cell potential is given by: E°cell = E°cathode - E°anode The standard reduction potential for the half-reaction Al³⁺ + 3 e- → Al(s) is: E°red = -E°ox where E°ox is the standard oxidation potential, which is the negative of the given value: E°ox = -(-1.66 V) = +1.66 V Therefore: E°red = -E°ox = -1.66 V The standard cell potential is: E°cell = E°cathode - E°anode = E°red - E°red = 0 V To calculate the non-standard cell potential, we use the Nernst equation: E = E°cell - (RT/nF) ln Q where: - R is the gas constant (8.314 J/(mol·K)) - T is the temperature in Kelvin (25 °C = 298 K) - n is the number of electrons transferred in the balanced half-reaction (3) - F is the Faraday constant (96485 C/mol) - Q is the reaction quotient, which for a concentration cell is the ratio of the concentrations of the reactants and products, each raised to their stoichiometric coefficients. In this case, Q is: Q = [Al³⁺(0.100 mol L^-1)] / [Al³⁺(1.0 × 10^-5 mol L^-1)] Q = 1.0 × 10^-6 Therefore, E = E°cell - (RT/nF) ln Q = 0 - (8.314 J/(mol·K) × 298 K / (3 × 96485 C/mol)) ln(1.0 × 10^-6) E = +0.0789 V Therefore, the answer is B: E° = 0.00 V and E = +0.0789 V.

Given That E°red = -1

Given That E°_red = -1