Calculate the standard cell voltage, Eo, for the following reactions. Include the sign of the voltage in your answers.
(a) Fe2+(aq) + Pb(s) --> Pb2+(aq) + Fe(s)
does charle's law depend on the identity of the gas
does charles law depend on the identity of the gas
If you wish to post a new question, go to the top of the page, click on post a new question and do so. Questions piggy-backed onto another post USUALLY get ignored. In this case I came back to look. No, Charles' Law does not depend upon the gas (as long as we are talking about ideal gas behavior). Few gases are ideal.
To calculate the standard cell voltage, Eo, for the given reaction, you need to use the standard reduction potentials of the half-reactions involved.
The given reaction involves two half-reactions:
1. Reduction of Fe2+(aq) to Fe(s): Fe2+(aq) + 2e- --> Fe(s)
2. Oxidation of Pb(s) to Pb2+(aq): Pb(s) --> Pb2+(aq) + 2e-
Step 1: Look up the standard reduction potentials (Eo) for each half-reaction. These values can be found in tables or reference books. Let's assume the values are:
- Eo(Fe2+(aq) + 2e- --> Fe(s)) = -0.44 V
- Eo(Pb(s) --> Pb2+(aq) + 2e-) = -0.13 V
Step 2: Identify the oxidizing agent and the reducing agent in the overall reaction. In this case, Fe2+(aq) is being reduced to Fe(s) (gaining electrons), so it is the reducing agent. Pb(s) is being oxidized to Pb2+(aq) (losing electrons), so it is the oxidizing agent.
Step 3: Assign the appropriate sign (+/-) to the reduction potentials based on the half-reactions. Since Fe2+(aq) is being reduced (gaining electrons), its reduction potential needs to be reversed. Pb(s) is being oxidized (losing electrons), so its reduction potential remains as given.
Step 4: Calculate the overall standard cell voltage (Eo) by adding the reduction potentials of the two half-reactions. The reduction potential of the reducing agent is multiplied by the stoichiometric coefficient, while the oxidizing agent's reduction potential remains unchanged.
Eo = Eo(Fe2+(aq) + 2e- --> Fe(s)) + Eo(Pb(s) --> Pb2+(aq) + 2e-)
= -0.44 V + (-0.13 V)
= -0.57 V
Therefore, the standard cell voltage, Eo, for the reaction Fe2+(aq) + Pb(s) --> Pb2+(aq) + Fe(s) is -0.57 V.