Two litre solution of a buffer mixture containing 1.0 M - NaH2PO4 and 1.0 M - Na2HPO4 is placed in t — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Total Electrical Charge and Moles of Electrons Transferred The total quantity of electrical charge ($Q$) passed through the electrolytic cell is determined 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 ($F = 96,500\text{ C mol}^{-1}$), the number of moles of electrons ($n_{e^-}$) passed during the electrolysis is: $$n_{e^-} = \frac{Q}{F}$$ $$n_{e^-} = \frac{72,375\text{ C}}{96,500\text{ C mol}^{-1}} = 0.75\text{ mol}$$ 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 and moles of the dihydrogen phosphate ion ($\ce{H2PO4^-}$, weak acid) and the hydrogen phosphate ion ($\ce{HPO4^{2-}}$, conjugate base) in each $1\text{ L}$ compartment are: * $$\text{Initial moles of }\ce{H2PO4^-} = 1.0\text{ M} \times 1\text{ L} = 1.0\text{ mol}$$ * $$\text{Initial moles of }\ce{HPO4^{2-}} = 1.0\text{ M} \times 1\text{ L} = 1.0\text{ mol}$$ Step 3 - Analyze the Anode Compartment and Calculate the Final pH At the anode, the oxidation of water takes place, which generates oxygen gas and hydrogen ions ($\ce{H^+}$): $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ According to this stoichiometry, $4\text{ moles}$ of electrons correspond to the production of $4\text{ moles}$ of $\ce{H^+}$ ions. Thus, the moles of $\ce{H^+}$ ions generated in the anode compartment is equal to the moles of electrons passed: $$n(\ce{H^+}) = 0.75\text{ mol}$$ These newly generated hydrogen ions ($\ce{H^+}$) react quantitatively with the basic component of our buffer ($\ce{HPO4^{2-}}$) to form $\ce{H2PO4^-}$ because the neutralization reaction goes virtually to completion: $$\ce{HPO4^{2-}(aq) + H^+(aq) -> H2PO4^-(aq)}$$ We calculate the final moles of each buffer component in the $1\text{ L}$ anode compartment: * $$\text{Final moles of conjugate base, }\ce{HPO4^{2-}} = 1.0\text{ mol} - 0.75\text{ mol} = 0.25\text{ mol}$$ * $$\text{Final moles of weak acid, }\ce{H2PO4^-} = 1.0\text{ mol} + 0.75\text{ mol} = 1.75\text{ mol}$$ 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 establishes 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 reduction of water takes place, which generates hydrogen gas and hydroxide ions ($\ce{OH^-}$): $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)}$$ According to this stoichiometry, $2\text{ moles}$ of electrons correspond to the production of $2\text{ moles}$ of $\ce{OH^-}$ ions. Thus, the moles of $\ce{OH^-}$ ions generated in the cathode compartment is equal to the moles of electrons passed: $$n(\ce{OH^-}) = 0.75\text{ mol}$$ These newly generated hydroxide ions ($\ce{OH^-}$) react quantitatively with the acidic component of our 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 buffer component in the $1\text{ L}$ cathode compartment: * $$\text{Final moles of weak acid, }\ce{H2PO4^-} = 1.0\text{ mol} - 0.75\text{ mol} = 0.25\text{ mol}$$ * $$\text{Final moles of conjugate base, }\ce{HPO4^{2-}} = 1.0\text{ mol} + 0.75\text{ mol} = 1.75\text{ mol}$$ 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 establishes 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). $$\text{Correct Options: } \boxed{B,C}$$