A voltaic cell utilizes the following reaction and operates at 298 K: $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 \mathrm{M},\) \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M},\) and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when $\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M}$ and \(\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M} ?\)

To calculate the emf of the cell under standard conditions, we will use the equation: \(E_{cell} = E_{cathode} - E_{anode}\) First, we need to look up the standard electrode potentials for the half-reactions. From the given equation, the half-cell reactions are: \[ \begin{aligned} &\text{Anode: } 3\,\mathrm{Ce^{3+}(aq)} \rightarrow 3\,\mathrm{Ce^{4+}(aq)} + 3\, e^- \quad (E^\circ_{\mathrm{Ce^{3+}/Ce^{4+}}}) \\ &\text{Cathode: } \mathrm{Cr^{3+}(aq)} + 3\, e^- \rightarrow \mathrm{Cr(s)} \quad (E^\circ_{\mathrm{Cr^{3+}/Cr}}) \end{aligned} \] Using a standard reduction potential table, we find: \(E^\circ_{\mathrm{Ce^{3+}/Ce^{4+}}} = 1.61 \,\mathrm{V}\) and \(E^\circ_{\mathrm{Cr^{3+}/Cr}} = -0.74\, \mathrm{V}\) Now, we can calculate the emf under standard conditions: \(E_{cell} = E_{cathode} - E_{anode} = (-0.74\, \mathrm{V}) - (1.61\, \mathrm{V}) = -2.35\, \mathrm{V}\) (a) The emf of the cell under standard conditions is -2.35 V.

We will now use the Nernst equation to find the emf of the cell under the given concentrations: \(E = E^\circ - \frac{RT}{nF} \ln Q\) where: - \(E\) is the cell potential - \(E^\circ\) is the standard cell potential - \(R\) is the universal gas constant - \(T\) is the temperature (in Kelvin) - \(n\) is the number of moles of electrons transferred in the reaction - \(F\) is Faraday's constant - \(Q\) is the reaction quotient For (b), the concentrations are as follows: \[\mathrm{[Ce^{4+}]}= 3.0\; \mathrm{M},\quad [\mathrm{Ce^{3+}}] = 0.10\; \mathrm{M},\quad [\mathrm{Cr^{3+}}] = 0.010\; \mathrm{M}\] The reaction quotient for this reaction is: \(Q = \frac{[\mathrm{Ce^{4+}]}^3[\mathrm{Cr^{3+}]} }{[\mathrm{Ce^{3+}]}^3} \) Plug in the given values and calculate Q: \(Q = \frac{(3.0)^3 \cdot (0.010)}{(0.10)^3} = 900\) Now calculate the emf for this condition: \(E = -2.35 - \frac{(8.314)(298)}{(3)(96485)} \ln (900)\) \(E = -2.35 - 0.0159 \ln(900) \approx -1.99\; \mathrm{V}\) (b) The emf of the cell with the given concentrations is approximately -1.99 V.

For (c), the concentrations are as follows: \(\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M} \text{ and }\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M}\) Calculate the reaction quotient for these concentrations: \(Q = \frac{(0.010)^3 \cdot (1.5)}{(2.0)^3} = 0.0009375\) Calculate the emf using the Nernst equation: \(E = -2.35 - \frac{(8.314)(298)}{(3)(96485)} \ln (0.0009375)\) \(E = -2.35 - 0.0159 \ln(0.0009375) \approx -2.51\; \mathrm{V}\) (c) The emf of the cell with the given concentrations is approximately -2.51 V.