This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions about the electrochemical cell: 8.1: The part labelled X is the salt bridge. • It completes the electrical circuit by allowing the movement of ions between the two half-cells. • It maintains electrical neutrality in the half-cells by preventing charge build-up. 8.2: To determine the anode, we compare the standard reduction potentials of Cu and Ag. Cu^2+(aq) + 2e^- Cu(s) E^ = +0.34 V Ag^+(aq) + e^- Ag(s) E^ = +0.80 V Silver has a higher reduction potential, meaning Ag^+ is more easily reduced (cathode). Copper has a lower reduction potential, meaning Cu is more easily oxidized (anode). The anode is Cu. 8.3: 8.3.1: The electrolyte in the Cu half-cell must contain Cu^2+ ions. The formula is CuSO_4 (or Cu(NO_3)_2). 8.3.2: The electrolyte in the Ag half-cell must contain Ag^+ ions. The formula is AgNO_3. 8.4: Step 1: Write the half-reactions. Anode (oxidation): Cu(s) Cu^2+(aq) + 2e^- Cathode (reduction): Ag^+(aq) + e^- Ag(s) Step 2: Balance the electrons by multiplying the cathode reaction by 2. 2Ag^+(aq) + 2e^- 2Ag(s) Step 3: Add the balanced half-reactions to get the net ionic equation. Cu(s) + 2Ag^+(aq) Cu^2+(aq) + 2Ag(s) The balanced net (overall) equation is Cu(s) + 2Ag^+(aq) Cu^2+(aq) + 2Ag(s). 8.5: Step 1: Identify the standard reduction potentials for the cathode and anode. E^_cathode = E^(Ag^+/Ag) = +0.80 V E^_anode = E^(Cu^2+/Cu) = +0.34 V Step 2: Calculate the standard cell potential using the formula E^_cell = E^_cathode - E^_anode. E^_cell = (+0.80 V) - (+0.34 V) E^_cell = 0.46 V The cell potential is +0.46 V. 8.6: The standard conditions under which the cell operates are: • Temperature of 25^ (298 K). • Concentration of 1 mol·dm^-3 (1 M) for all aqueous solutions. Drop the next question!