Find the emf for the cell
Cr(s) |Cr3+(aq, 0.37 M) ||
Pb2+(aq, 0.0095 M) | Pb(s).
Answer in units of V
I got .548733V
Is this right?
wait nvm, i got .55875
no
i got .57876
.55875 is correct
Hey there! Well, if you've gotten an answer of .548733V, then it seems like you've really put a lot of effort into crunching those numbers. However, I must apologize because your answer is not quite right. The correct emf for the cell in this case is actually 0.136V. Remember, it's all about that positive thinking... and the right calculations, of course! Keep up the good work!
To find the electromotive force (emf) for a cell, you need to use the standard reduction potentials of the half-reactions involved. The emf for a cell is the difference between the reduction potentials of the two half-reactions.
In this case, we have two half-reactions:
Cr(s) + 3e- → Cr3+(aq)
Pb2+(aq) + 2e- → Pb(s)
Here are the standard reduction potentials (E°) for these reactions:
Cr3+(aq) + 3e- → Cr(s): E° = -0.74 V
Pb2+(aq) + 2e- → Pb(s): E° = -0.13 V
To calculate the emf of the cell, you need to subtract the reduction potential of the anode half-reaction (where oxidation occurs) from the reduction potential of the cathode half-reaction (where reduction occurs):
Emf = E°(cathode) - E°(anode)
In this case, Pb2+(aq) + 2e- → Pb(s) is the cathode half-reaction, so its reduction potential is -0.13 V. Cr(s) + 3e- → Cr3+(aq) is the anode half-reaction, so its reduction potential is -0.74 V.
Emf = (-0.13 V) - (-0.74 V)
Emf = 0.61 V
Therefore, the answer for the emf of the cell in units of volts is 0.61 V, not 0.548733 V as you calculated.