Two litre solution of buffer 1.0 M-NaH2PO4 and 1.0 M-Na2HPO4 in two compartments. 1.25 A for 965 min — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Total Electric Charge and Moles of Electrons Transferred The total electrical charge ($Q$) passed through the cell is calculated using the formula: $$Q = I \times t$$ Where: * Current ($I$) = $1.25\text{ A}$ * Time ($t$) = $965\text{ min} = 965 \times 60\text{ s} = 57,900\text{ s}$ Substituting the given values into the formula: $$Q = 1.25\text{ A} \times 57,900\text{ s} = 72,375\text{ C}$$ Using Faraday's constant ($F = 96,500\text{ C mol}^{-1}$), we calculate the total moles of electrons ($n_{e^-}$) passed during the electrolysis: $$n_{e^-} = \frac{Q}{F}$$ $$n_{e^-} = \frac{72,375\text{ C}}{96,500\text{ C mol}^{-1}} = 0.75\text{ mol}$$ Step 2 - Determine the Initial Moles of Buffer Components in Each Compartment The total volume of the buffer solution is $2\text{ L}$, which is divided equally into two separate compartments (anode and cathode). Therefore, each compartment contains exactly $1\text{ L}$ of the buffer solution. The initial concentrations of the buffer components in each $1\text{ L}$ compartment are: * Dihydrogen phosphate ion ($\ce{H2PO4^-}$): $[\ce{H2PO4^-}]_0 = 1.0\text{ M} \implies n_0(\ce{H2PO4^-}) = 1.0\text{ mol}$ * Hydrogen phosphate ion ($\ce{HPO4^{2-}}$): $[\ce{HPO4^{2-}}]_0 = 1.0\text{ M} \implies n_0(\ce{HPO4^{2-}}) = 1.0\text{ mol}$ Step 3 - Analyze the Anode Compartment and Calculate the Final pH At the anode, the electrolysis of water takes place to release oxygen gas and hydrogen ions ($\ce{H^+}$): $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ According to this half-reaction, $4\text{ moles}$ of electrons produce $4\text{ moles}$ of $\ce{H^+}$ ions. Thus, the moles of $\ce{H^+}$ ions generated at the anode is equal to the moles of electrons passed: $$n(\ce{H^+}) = n_{e^-} = 0.75\text{ mol}$$ These newly generated hydrogen ions ($\ce{H^+}$) react quantitatively with the basic component of the buffer ($\ce{HPO4^{2-}}$) to form $\ce{H2PO4^-}$ because the reaction goes virtually to completion: $$\ce{HPO4^{2-}(aq) + H^+(aq) -> H2PO4^-(aq)}$$ We calculate the final moles of each species in the $1\text{ L}$ anode compartment: * Final moles of conjugate base, $\ce{HPO4^{2-}}$: $$n(\ce{HPO4^{2-}})_{\text{final}} = n_0(\ce{HPO4^{2-}}) - n(\ce{H^+}) = 1.0\text{ mol} - 0.75\text{ mol} = 0.25\text{ mol}$$ * Final moles of weak acid, $\ce{H2PO4^-}$: $$n(\ce{H2PO4^-})_{\text{final}} = n_0(\ce{H2PO4^-}) + n(\ce{H^+}) = 1.0\text{ mol} + 0.75\text{ mol} = 1.75\text{ mol}$$ Applying the Henderson-Hasselbalch equation for the acid buffer system: $$\text{pH}_{\text{anode}} = \text{p}K_a + \log_{10}\left(\frac{[\ce{HPO4^{2-}}]}{[\ce{H2PO4^-}]}\right)$$ $$\text{pH}_{\text{anode}} = 2.15 + \log_{10}\left(\frac{0.25\text{ mol / 1 L}}{1.75\text{ mol / 1 L}}\right)$$ $$\text{pH}_{\text{anode}} = 2.15 + \log_{10}\left(\frac{1}{7}\right)$$ $$\text{pH}_{\text{anode}} = 2.15 - \log_{10}(7)$$ Substituting the given value of $\log_{10}(7) = 0.85$: $$\text{pH}_{\text{anode}} = 2.15 - 0.85 = \boxed{1.30}$$ This calculation confirms that **Option (C) is correct** and Option (A) is incorrect. Step 4 - Analyze the Cathode Compartment and Calculate the Final pH At the cathode, the electrolysis of water takes place to release hydrogen gas and hydroxide ions ($\ce{OH^-}$): $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)}$$ According to this half-reaction, $2\text{ moles}$ of electrons produce $2\text{ moles}$ of $\ce{OH^-}$ ions. Thus, the moles of $\ce{OH^-}$ ions generated at the cathode is equal to the moles of electrons passed: $$n(\ce{OH^-}) = n_{e^-} = 0.75\text{ mol}$$ These newly generated hydroxide ions ($\ce{OH^-}$) react quantitatively with the acidic component of the buffer ($\ce{H2PO4^-}$) to form $\ce{HPO4^{2-}}$ and water: $$\ce{H2PO4^-(aq) + OH^-(aq) -> HPO4^{2-}(aq) + H2O(l)}$$ We calculate the final moles of each species in the $1\text{ L}$ cathode compartment: * Final moles of weak acid, $\ce{H2PO4^-}$: $$n(\ce{H2PO4^-})_{\text{final}} = n_0(\ce{H2PO4^-}) - n(\ce{OH^-}) = 1.0\text{ mol} - 0.75\text{ mol} = 0.25\text{ mol}$$ * Final moles of conjugate base, $\ce{HPO4^{2-}}$: $$n(\ce{HPO4^{2-}})_{\text{final}} = n_0(\ce{HPO4^{2-}}) + n(\ce{OH^-}) = 1.0\text{ mol} + 0.75\text{ mol} = 1.75\text{ mol}$$ Applying the Henderson-Hasselbalch equation for the cathode compartment: $$\text{pH}_{\text{cathode}} = \text{p}K_a + \log_{10}\left(\frac{[\ce{HPO4^{2-}}]}{[\ce{H2PO4^-}]}\right)$$ $$\text{pH}_{\text{cathode}} = 2.15 + \log_{10}\left(\frac{1.75\text{ mol / 1 L}}{0.25\text{ mol / 1 L}}\right)$$ $$\text{pH}_{\text{cathode}} = 2.15 + \log_{10}(7)$$ Substituting the given value of $\log_{10}(7) = 0.85$: $$\text{pH}_{\text{cathode}} = 2.15 + 0.85 = \boxed{3.00}$$ This calculation confirms that **Option (B) is correct** and Option (D) is incorrect. Step 5 - Conclusion Evaluating the calculations: * The final pH at the anode is $1.30$. * The final pH at the cathode is $3.00$. Therefore, the correct options are (B) and (C). $$\text{Correct Options: } \boxed{B,C}$$