Could I get some help with these please?
Thanks
Calculate the electrical potential (V).
Ag/1M Ag+1// 1M Fe+3/Fe
Ag2+( aq ) + e – Ag+( aq ) 1.98
Fe3+( aq ) + 3e – Fe(s) –0.04
2. A voltaic cell is reconstructed used electrodes based on the following half reactions:
Pb2+ ( aq ) + 2e – Pb (s) -0.13V
Mn2+ ( aq ) + 2e – Mn (s) -1.19V
a. What is the anode and what is the cathode in this cell?
b. What is the standard cell potential?
Pb2+ ( aq ) + 2e – Pb (s) -0.13V
Mn2+ ( aq ) + 2e – Mn (s) -1.19V
Pb2+ ( aq ) + 2e – Pb (s) -0.13V
Mn (s) Mn2+ ( aq ) + 2e – +1.19V
This doesn't look like help. It looks like we're doing your homework. Tell us what you don't understand about the first one and we can help you through it.
Sure! I'd be happy to help you with these questions. Let's go through them step by step.
1. Calculate the electrical potential (V) for the given cell:
Ag/1M Ag+1// 1M Fe+3/Fe
The given half-reactions are:
Ag2+( aq ) + e – Ag+( aq ) E° = 1.98 V
Fe3+( aq ) + 3e – Fe(s) E° = -0.04 V
To calculate the overall cell potential, we need to add the reduction potentials of the half-reactions. However, to do that, we need to make sure the reactions are balanced.
Since we are missing the reduction potential for the Ag+1 to Ag couple, we can use the Nernst equation to calculate it. The Nernst equation is given as:
E = E° - (RT/nF) * ln(Q)
In this case, the equation becomes:
E = E° - (0.0257/n) * ln(Q)
We know that Q is the reaction quotient and n is the number of electrons transferred in the half-reaction.
Now, let's calculate the reduction potential for the Ag+1 to Ag couple:
Q = [Ag+1] / [Ag] = [1M] / [1M] = 1
n = 1 (since 1 electron is transferred)
Substituting these values into the Nernst equation:
E = 1.98 - (0.0257/1) * ln(1)
E = 1.98 - 0
E = 1.98 V
Now that we have the reduction potential for the Ag+1 to Ag couple, we can calculate the overall cell potential:
Ecell = E(cathode) - E(anode)
Ecell = E(Fe3+ to Fe) - E(Ag+1 to Ag)
Ecell = -0.04 - 1.98
Ecell = -2.02 V
Therefore, the electrical potential (V) for the given cell is -2.02 V.
2. Now let's move on to the second question about a voltaic cell with Pb2+ and Mn2+ half-reactions.
a. To find the anode and cathode, we need to compare the reduction potentials of the half-reactions. The half-reaction with the more positive reduction potential is the cathode, and the one with the less positive reduction potential is the anode.
Comparing the reduction potentials:
Pb2+ ( aq ) + 2e – Pb (s) E° = -0.13 V
Mn2+ ( aq ) + 2e – Mn (s) E° = -1.19 V
Since the reduction potential for Pb2+ to Pb (-0.13 V) is more positive than that for Mn2+ to Mn (-1.19 V), the cathode is Pb2+ ( aq ) + 2e – Pb (s) and the anode is Mn (s) Mn2+ ( aq ) + 2e – .
b. To find the standard cell potential, we simply subtract the reduction potential of the anode from the reduction potential of the cathode:
Standard cell potential (E°cell) = E(cathode) - E(anode)
E°cell = -0.13 - (-1.19)
E°cell = 1.06 V
Therefore, the standard cell potential for the given cell is 1.06 V.
I hope this helps! Let me know if you have any further questions.