Calculate the voltage produced by the cell Sn(s)|Sn2+|Ag(s) at 25°C given: [Sn2+]=0.15 M [Ag+]=1.7 M

Step 1: Identify the half-cells and their roles. The galvanic cell consists of a hydrogen half-cell and a magnesium half-cell. The standard reduction potential for the hydrogen half-cell (Standard Hydrogen Electrode, SHE) is defined as 0.00 \, V. 2H^+(aq) + 2e^- H_2(g) E^ = 0.00 \, V Magnesium is a more reactive metal than hydrogen, indicating it has a greater tendency to be oxidized. Therefore, the magnesium electrode will act as the anode (where oxidation occurs), and the hydrogen electrode will act as the cathode (where reduction occurs). Step 2: Write the half-reactions for the anode and cathode. Anode (oxidation): Mg(s) Mg^2+(aq) + 2e^- Cathode (reduction): 2H^+(aq) + 2e^- H_2(g) Step 3: Write the overall cell reaction. Adding the anode and cathode half-reactions: Mg(s) + 2H^+(aq) Mg^2+(aq) + H_2(g) Step 4: Calculate the standard reduction potential for the magnesium half-cell. The cell potential (E_cell) is given as 2.36 V. The formula for the standard cell potential is: E^_cell = E^_cathode - E^_anode Substituting the known values: 2.36 \, V = 0.00 \, V - E^_Mg^2+/Mg Solving for E^_Mg^2+/Mg: E^_Mg^2+/Mg = -2.36 \, V Send me the next one 📸