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
Can someone please help me?
I have the equation:
2Cr +3Pb(+2) --> 2Cr(+3) + 3Pb
See your post above.
To find the electromotive force (emf) for the given cell, we can use the standard reduction potentials of the half-reactions involved. The standard reduction potential is a measure of the tendency of a species to gain electrons and undergo reduction.
First, let's write the balanced half-reactions for the two electrodes involved:
1. For the anode (oxidation half-reaction at the Cr electrode):
2Cr(s) → 2Cr3+(aq) + 6e-
2. For the cathode (reduction half-reaction at the Pb electrode):
3Pb2+(aq) + 6e- → 3Pb(s)
The emf of the cell (Ecell) can be calculated using the formula: Ecell = E(cathode) - E(anode)
Now, we need to look up the standard reduction potentials (E°) for the Cr3+(aq)/Cr(s) and Pb2+(aq)/Pb(s) half-reactions. This information is usually available in electrochemical tables or reference books.
From the table, we find that the standard reduction potential of the Cr3+(aq)/Cr(s) half-reaction is +0.74 V, and the standard reduction potential for the Pb2+(aq)/Pb(s) half-reaction is -0.13 V.
Now, substitute these values into the formula to calculate the emf of the cell:
Ecell = E(cathode) - E(anode)
Ecell = -0.13 V - (+0.74 V)
Ecell = -0.13 V - 0.74 V
Ecell = -0.87 V
Therefore, the emf of the cell is -0.87 V. Note that the negative sign indicates that the reaction is not spontaneous in the forward direction by itself.