Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the direction of electron flow.
b. 2 H⁺(aq) + Fe(s) → H₂(g) + Fe²⁺(aq)
c. 2 NO₃^–(aq) + 8 H⁺(aq) + 3 Cu(s) → 2 NO(g) + 4 H₂O(l) + 3 Cu²⁺(aq)

Verified step by step guidance

1. Step 1: Identify the oxidation and reduction half-reactions for each redox reaction. For reaction b, Fe(s) is oxidized to Fe^{2+}(aq), and H^{+}(aq) is reduced to H_{2}(g). For reaction c, Cu(s) is oxidized to Cu^{2+}(aq), and NO_{3}^{-}(aq) is reduced to NO(g).
2. Step 2: Determine the anode and cathode for each reaction. The anode is where oxidation occurs, and the cathode is where reduction occurs. For reaction b, Fe(s) is the anode, and H^{+}(aq) is the cathode. For reaction c, Cu(s) is the anode, and NO_{3}^{-}(aq) is the cathode.
3. Step 3: Write the half-reactions for each electrode. For reaction b, the anode half-reaction is Fe(s) → Fe^{2+}(aq) + 2e^{-}, and the cathode half-reaction is 2H^{+}(aq) + 2e^{-} → H_{2}(g). For reaction c, the anode half-reaction is Cu(s) → Cu^{2+}(aq) + 2e^{-}, and the cathode half-reaction is 2NO_{3}^{-}(aq) + 8H^{+}(aq) + 6e^{-} → 2NO(g) + 4H_{2}O(l).
4. Step 4: Label the species present in each solution. For reaction b, the anode solution contains Fe^{2+}(aq), and the cathode solution contains H^{+}(aq). For reaction c, the anode solution contains Cu^{2+}(aq), and the cathode solution contains NO_{3}^{-}(aq) and H^{+}(aq).
5. Step 5: Indicate the direction of electron flow. Electrons flow from the anode to the cathode in both reactions. For reaction b, electrons flow from Fe(s) to H^{+}(aq). For reaction c, electrons flow from Cu(s) to NO_{3}^{-}(aq).