91Ó°ÊÓ

To write the balanced chemical equation for the reduction of cassiterite with carbon, we need to consider the reactants involved and the products formed. Cassiterite is SnO_2 and carbon (in the form of coke) is used as a reducing agent. The products formed are tin (Sn) and carbon dioxide (CO_2). The unbalanced equation would be: SnO_2 + C → Sn + CO_2 Now, we need to balance the equation: \(2\text{SnO}_2 + 2\text{C} \rightarrow 2\text{Sn} + 2\text{CO}_2 \)

In the electrolysis of SnSO_4, the following reactions occur: 1. At the cathode (reduction): Sn^(2+) + 2e^- → Sn This reaction represents tin ions accepting electrons to become tin metal. 2. At the anode (oxidation): Since the solution is acidic, the water molecules will be oxidized to oxygen gas, and hydrogen ions will be left behind in the solution: 2H_2O → O_2 + 4H^+ + 4e^- This reaction represents the oxidation of water molecules producing oxygen gas, hydrogen ions, and releasing electrons. Now that we have the electrode reactions, we can write the balanced equation for the whole electrolysis process by combining the cathode and anode reactions.

To combine the cathode and anode reactions, we need to balance the number of electrons involved in both reactions. In the cathode reaction, 2 electrons are involved, while in the anode reaction, 4 electrons are involved. So, we need to multiply the cathode reaction by 2 to balance the electrons: Cathode reaction (multiplied by 2): 2Sn^(2+) + 4e^- → 2Sn Anode reaction: 2H_2O → O_2 + 4H^+ + 4e^- Balanced electrolysis equation: 2Sn^(2+) + 2H_2O → 2Sn + O_2 + 4H^+ Therefore, the balanced chemical equations for the reduction process and the electrode reactions are: Reduction of cassiterite with carbon: \(2\text{SnO}_2 + 2\text{C} \rightarrow 2\text{Sn} + 2\text{CO}_2 \) Electrolysis of SnSO_4 in an acidic solution: 2Sn^(2+) + 2H_2O → 2Sn + O_2 + 4H^+