A voltaic cell utilizes the following reaction and operates at 298 K: 3 Ce⁴⁺(aq) + Cr(s) → 3 Ce³⁺(aq) + Cr³⁺(aq) (b) What is the emf of this cell when [Ce⁴⁺] = 3.0 M, [Ce³⁺] = 0.10 M, and [Cr³⁺] = 0.010 M? (c) What is the emf of the cell when [Ce⁴⁺] = 0.010 M, [Ce³⁺] = 2.0 M, and [Cr³⁺] = 1.5 M?

Verified step by step guidance

1. Step 1: Identify the half-reactions and their standard reduction potentials. The given reaction is 3 Ce⁴⁺(aq) + Cr(s) → 3 Ce³⁺(aq) + Cr³⁺(aq). The half-reactions are: Ce⁴⁺ + e⁻ → Ce³⁺ and Cr → Cr³⁺ + 3e⁻. Look up the standard reduction potentials (E°) for these half-reactions in a standard reduction potential table.
2. Step 2: Calculate the standard cell potential (E°cell). Use the formula E°cell = E°cathode - E°anode. The cathode is where reduction occurs (Ce⁴⁺ to Ce³⁺), and the anode is where oxidation occurs (Cr to Cr³⁺). Substitute the standard reduction potentials into the formula to find E°cell.
3. Step 3: Use the Nernst equation to calculate the cell potential (Ecell) under non-standard conditions. The Nernst equation is Ecell = E°cell - (RT/nF) * ln(Q), where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), n is the number of moles of electrons transferred (3 in this case), F is Faraday's constant (96485 C/mol), and Q is the reaction quotient.
4. Step 4: Calculate the reaction quotient (Q) for the given concentrations. For the reaction 3 Ce⁴⁺(aq) + Cr(s) → 3 Ce³⁺(aq) + Cr³⁺(aq), Q = ([Ce³⁺]^3 * [Cr³⁺]) / [Ce⁴⁺]^3. Substitute the given concentrations into this expression to find Q for each scenario.
5. Step 5: Substitute the values of E°cell, R, T, n, F, and Q into the Nernst equation to calculate the emf of the cell for each set of conditions. This will give you the emf for both scenarios: (b) [Ce⁴⁺] = 3.0 M, [Ce³⁺] = 0.10 M, [Cr³⁺] = 0.010 M and (c) [Ce⁴⁺] = 0.010 M, [Ce³⁺] = 2.0 M, [Cr³⁺] = 1.5 M.