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.
ScanSolve AI Answer
Here is the answer for question 28 b): b) Step 1: Compare the standard electrode potentials (E^) of copper and Q. E^_Cu^2+/Cu = +0.34 V E^_Q^+/Q = +0.80 V Step 2: Determine the relative reactivity. Since E^_Cu^2+/Cu is less positive than E^_Q^+/Q, copper metal is a stronger reducing agent (more easily oxidized) than metal Q. Step 3: Explain the reaction. Because copper is a stronger reducing agent, it will spontaneously reduce the Q^+ ions in the solution to solid Q, while the copper metal itself gets oxidized to Cu^2+ ions. This means copper displaces Q from its nitrate solution. Reason: Copper is a stronger reducing agent (more reactive) than metal Q. Therefore, copper metal is oxidized to Cu^2+ ions, causing the copper strip to dissolve, while Q^+ ions are reduced to solid Q, which deposits on the copper strip. Last free one today — make it count tomorrow, or type /upgrade for unlimited.