Question

Given the following information: $$ \begin{array}{ll} \mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{CrO}_{4}^{2-}(a q) & K_{\mathrm{sp}}=1 \times 10^{-12} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.799 \mathrm{~V} \end{array} $$ find the standard reduction potential at \(25^{\circ} \mathrm{C}\) for the half- reaction $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)+2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{CrO}_{4}^{2-}(a q) $$

Short answer

Answer: The standard reduction potential for the given half-reaction at 25°C is approximately 0.489 V.

Step-by-step solution

Write the Nernst equation for silver ion reduction half-cell

The Nernst equation relates the cell potential (E) to the standard cell potential (E°), temperature (T), number of moles of electrons (n), the gas constant (R), and the Faraday constant (F). It is given by: $$ E = E^{\circ} - \frac{RT}{nF} \ln Q $$ Since we are working at standard conditions and only need the standard reduction potential, Q = 1, ln Q = 0, and the Nernst equation simplifies to: $$ E^{\circ} = E $$ Since we are given E° for the reduction of silver ions, we can use this directly for the first half-reaction.

Relate Ksp and E° for the dissolution of silver chromate

We are given Ksp for the dissolution of silver chromate: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{CrO}_{4}^{2-}(a q), \; K_{\mathrm{sp}}=1 \times 10^{-12} $$ We can relate this to the standard cell potential (E°) using the relationship between equilibrium constants and cell potentials: $$ E^{\circ} = -\frac{RT}{nF} \ln K $$ Since this is an equilibrium between Ag2CrO4 and dissolved Ag+ and CrO42-, n = 2 (two electrons are exchanged). At 25°C, we can plug in the values of R, T, F, and n to find E° for this reaction.

Calculate the E° for dissolving silver chromate

Using the equation from Step 2, we can solve for the standard cell potential for dissolving silver chromate: $$ E^{\circ} = -\frac{8.314 \: J/(mol \cdot K) \times 298 \: K}{2 \times 96,485 \: C/mol} \ln(1 \times 10^{-12}) $$ $$ E^{\circ} \approx -0.31 \: V $$ This is the standard cell potential for the dissolution of Ag2CrO4.

Add the two half-cell potentials to find the standard reduction potential for the given half-reaction

To find the standard reduction potential for the given half-reaction: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)+2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{CrO}_{4}^{2-}(a q), $$ we add the standard reduction potential for Ag+ reduction and the standard cell potential for Ag2CrO4 dissolution: $$ E^{\circ}_{Ag_{2}CrO_{4}/Ag} = E^{circ}_{Ag^{+}/Ag} + E^{\circ}_{Ag_{2}CrO_{4(s)}} $$ $$ E^{\circ}_{Ag_{2}CrO_{4}/Ag} = 0.799 \: V + (-0.31 \: V) $$ $$ E^{\circ}_{Ag_{2}CrO_{4}/Ag} \approx 0.489 \: V $$ So, the standard reduction potential for the given half-reaction at 25°C is approximately 0.489 V.