A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when $\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\( \)96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\( and the \)\mathrm{pH}\( in both half-cells is \)4.00 ?$

In the given redox equation, the half-reactions are: - Fe³⁺(aq) gains an electron and is reduced to Fe²⁺(aq). This is the reduction half-reaction: $$\mathrm{Fe}^{3+}(a q)+1 e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)$$ - H₂(g) loses an electron and is oxidized into H⁺(aq). This is the oxidation half-reaction: $$\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q) + 2 e^{-}$$

To find the E°cell under standard conditions, we need to look up the standard reduction potentials for the two half-reactions. In this case, E°(Fe ³⁺/Fe²⁺)=+0.77 V, and E°(H⁺/H₂)=0.0 V for the standard hydrogen electrode (SHE). Now, add the two standard reduction potentials to find the standard cell potential. $$E°_{\text{cell}} = E°_{\text{Reduction}}+ E°_{\text{Oxidation}}$$ $$E°_{\text{cell}}= 0.77 \text{ V} + 0.0 \text{ V}$$ $$E°_{\text{cell}}= 0.77 \text{ V}$$ Thus, the emf of this cell under standard conditions is \(0.77 V\).

For part (b), we need to find the emf when we have the given concentrations and pressures. We’ll use the Nernst equation: \(E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF} \ln Q\), Here, E°cell is the standard cell potential, R is the gas constant (8.314 J/mol K), T is the temperature (assumed to be 298 K if not given), n is the number of electrons transferred, F is Faraday's constant (96485 C/mol), and Q is the reaction quotient for the given concentrations and pressures. First, find the number of electrons transferred, n. In the balanced reaction, the oxidation half-reaction involves 2 moles of electrons and the reduction half-reaction involves 1 mole of electrons. Thus, n = 2. Next, calculate the reaction quotient, Q. In this case, we have: $$Q= \frac{[\text{Fe}^{2+}]^2 [\text{H}^{+}]^2}{[\text{Fe}^{3+}]^2 P_{\text{H}_{2}}}$$ Since we are given the concentrations and pressure values, plug in those values: $$Q = \frac{(0.0010 \text{ M})^2 \cdot (10^{-4.00})^2}{(3.50 \text{ M})^2 * (96.3/101.325)}$$ Now, plug all the values into the Nernst equation: $$E_{\text{cell}} = 0.77 \text{ V} - \frac{8.314 \text{ J/mol K} \cdot 298 \text{ K}}{2 * 96485 \text{ C/mol}} \ln Q$$ Calculate the value of Ecell. Thus, the emf values for the voltaic cell (a) under standard conditions and (b) under the given conditions have been determined.