Question

PROBLEM SOLVING: Show pertinent solution and underline/put in a box your final answer. Use the space provided. 15 pts

1. Balance the ff skeleton reactions and identify the oxidizing and reducing agents: 6 pts
a) O2(g) + NO(g) + H2O(l) → NO3-(aq) + H+(aq)
b) CrO42–(aq) + H2O(l) + Cu(s) → Cr(OH)3(s) + Cu(OH)2(s) + OH–(aq)

2. A cell can be prepared from copper and tin. What is the E°cell for the cell that forms from the
following half-reactions? 3 pts
Cu2+(aq) + 2e- Cu(s); E° = 0.34 V Sn4+(aq) + 2e- Sn2+(aq); E° = 0.13 V

Answer 1

Based on the redox equations;

• O2 is the oxidizing agent while NO is the reducing agent
• Cu is the reducing agent while CrO_4{^2} is the oxidizing agent

The emf of the cell, E°cell of the cells formed from the half-reactions is 0.21 V

What are reducing and oxidizing agents?

Reducing agents are substances which donates electrons and are oxidized in a redox reaction.

Oxidizing agents are substances which accept electrons and are reduced in a redox reaction.

From the given the redox equations:

a) O_2(g) + NO(g) + H_2O(l) → NO_3^{-}(aq) + H^{+}(aq)

• O_2 is the oxidizing agent while NO is the reducing agent.

b) CrO_4^{2-}(aq) + H_2O(l) + Cu(s) → Cr(OH)_3(s) + Cu(OH)_2(s) + OH^{-}(aq)

• Cu is the reducing agent while CrO_4^{2-} is the oxidizing agent.

2. The emf of the cell, E°cell = E°red - E°oxi

In the cell, copper is reduced while tin is oxidized.

E°cell = (0.34 - 0.13) V

E°cell = 0.21 V

Therefore, the emf of the cell, E°cell is 0.21 V

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