Assuming that hydrogen behaves as an ideal gas, what is the EMF of the cell at 25°C if P1 = 600 mm a — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Half-Cell Reactions and Net Cell Reaction The cell is represented as: $$\text{Pt} \mid \ce{H2} (P_1) \mid \ce{HCl} \mid \ce{H2} (P_2) \mid \text{Pt}$$ This is a gas electrode concentration cell. We can identify the half-reactions at the electrodes as follows: * **At the Anode (Left - Oxidation):** $$\ce{H2}(P_1) \longrightarrow \ce{2H^+(aq)} + \ce{2e^-}$$ * **At the Cathode (Right - Reduction):** $$\ce{2H^+(aq)} + \ce{2e^-} \longrightarrow \ce{H2}(P_2)$$ Since both electrodes are in contact with the same $\ce{HCl}$ electrolyte solution, the concentration of hydrogen ions ($\ce{H^+}$) is identical at both half-cells. By adding the two half-reactions, we obtain the overall cell reaction: $$\ce{H2}(P_1) \longrightarrow \ce{H2}(P_2)$$ The number of moles of electrons transferred per mole of reaction is: $$n = 2$$ Step 2 - State the Nernst Equation for the Concentration Cell For any concentration cell where both electrodes are of the same type, the standard cell potential is exactly zero: $$E^\circ_{\text{cell}} = 0\text{ V}$$ Using the Nernst equation, the cell electromotive force ($E_{\text{cell}}$) at $25^\circ\text{C}$ is given by: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 R T}{n F} \log Q$$ Under the standard cell potential convention for gas concentration cells, the cell EMF can be written as: $$E_{\text{cell}} = \frac{2.303 R T}{n F} \log\left(\frac{P_2}{P_1}\right)$$ Substitute the given value $\frac{2.303 R T}{F} = 0.06\text{ V}$ and $n = 2$: $$E_{\text{cell}} = \frac{0.06\text{ V}}{2} \log\left(\frac{P_2}{P_1}\right)$$ $$E_{\text{cell}} = 0.03 \log\left(\frac{P_2}{P_1}\right)\text{ V}$$ Step 3 - Substitute Values and Calculate the EMF of the Cell We are given the following values: * $P_1 = 600\text{ mm}$ * $P_2 = 420\text{ mm}$ * $\log 7 = 0.85$ Substitute the pressures into our EMF equation: $$E_{\text{cell}} = 0.03 \log\left(\frac{420\text{ mm}}{600\text{ mm}}\right)\text{ V}$$ $$E_{\text{cell}} = 0.03 \log(0.7)\text{ V}$$ Now, calculate the logarithm of $0.7$ using the properties of logarithms: $$\log(0.7) = \log\left(\frac{7}{10}\right) = \log 7 - \log 10$$ Since $\log 10 = 1$ and $\log 7 = 0.85$: $$\log(0.7) = 0.85 - 1.0 = -0.15$$ Now, substitute this value back into the EMF equation: $$E_{\text{cell}} = 0.03 \times (-0.15)\text{ V}$$ $$E_{\text{cell}} = \boxed{-0.0045\text{ V}}$$ Step 4 - Evaluate the Options * **Option (A) is correct:** As mathematically calculated, the EMF of the concentration cell under these conditions is exactly $-0.0045\text{ V}$. * **Option (B) is incorrect:** $+0.0045\text{ V}$ is obtained if the negative sign convention in the Nernst equation is reversed, which corresponds to the spontaneous process direction rather than the standard representation of the written cell diagram. * **Option (C) is incorrect:** $+0.051\text{ V}$ represents a mathematical error where the logarithmic calculations or the number of transferred electrons ($n$) are misapplied. * **Option (D) is incorrect:** $-0.051\text{ V}$ represents a similar calculation and sign error. $$\text{Correct Option: } \boxed{\text{A}}$$