Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s)∣ Ni²⁺(aq,1.0 × 10^-5 M)∣∣ Ni²⁺(aq,0.100 M)∣ Ni(s) A)E° = 0.00 V and E = +0.24 V B)E° = 0.00 V and E = +0.12 V C)E° = -0.26 V and E = -0.02 V D)E° = -0.26 V and E = -0.14 V

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The shorthand notation provided in the problem is as follows: Ni(s) ∣ Ni²⁺(aq, 1.0 × 10^-5 M) ∣∣ Ni²⁺(aq, 0.100 M) ∣ Ni(s) To find E°, we can use the given standard reduction potential: E°_red = -0.26 V E° = E°(reduction at cathode) - E°(reduction at anode) = 0 - (-0.26 V) = +0.26 V To find E, we can use the equation: E = E° - (0.0592/n) log(Q) where n is the number of electrons transferred in the reaction and Q is the reaction quotient. In this case, the cell reaction is: Ni²⁺ + 2e- → Ni + 2^2- So n = 2. At equilibrium, the reaction quotient Q is: Q = [Ni2+(aq)]/[S2-(aq)]2 Since the concentrations of Ni2+ and S2- are the same on both sides of the cell, Q = 1. So, E = E° - (0.0592/n) log(Q) = E° - (0.0592/2) log(1) = E° = +0.26 V Therefore, the final answer is: Choice B) E° = 0.00 V and E = +0.12 V, as E° = +0.26 V and E = E° = +0.26 V - (0.0592/2) log(1) = +0.12 V.

Given That E°red = -0

Given That E°_red = -0