Given:
Cr3+(aq) + 3e¨C  Cr(s); E¡ã = ¨C0.74 V
Pb2+(aq) + 2e¨C  Pb(s); E¡ã = ¨C0.13 V
What is the standard cell potential for the following reaction?
2Cr(s) + 3Pb2+(aq) ¡ú 3Pb(s) + 2Cr3+(aq)
No arrows leaves me cold.
I can't translate some of the symbols.
Anyway, the answer to your problem is to add the oxidation volts to the reduction volts.
To find the standard cell potential for the given reaction, we need to sum up the standard reduction potentials of the half-reactions involved.
The standard cell potential, E°cell, is calculated using the equation: E°cell = E°cathode - E°anode, where E°cathode is the reduction potential of the cathode and E°anode is the reduction potential of the anode.
For the given reaction:
2Cr(s) + 3Pb2+(aq) → 3Pb(s) + 2Cr3+(aq)
The reduction half-reactions are:
Cr3+(aq) + 3e- → Cr(s) (E° = -0.74 V)
Pb2+(aq) + 2e- → Pb(s) (E° = -0.13 V)
To calculate the overall standard cell potential, we need to flip the sign of the reduction potential for the Cr3+(aq) → Cr(s) half-reaction since it's the oxidation half-reaction in the given reaction.
Flipping the sign gives us:
Cr(s) → Cr3+(aq) + 3e- (E° = +0.74 V)
Now, we can calculate the standard cell potential:
E°cell = E°cathode - E°anode
= (-0.13 V) - (+0.74 V)
E°cell = -0.87 V
Therefore, the standard cell potential for the given reaction is -0.87 V.
To find the standard cell potential for the given reaction, you need to use the standard reduction potentials of the half-reactions involved.
First, let's write the reduction half-reactions and their respective reduction potentials:
Reduction half-reaction 1:
Cr3+(aq) + 3e- --> Cr(s) (E° = -0.74 V)
Reduction half-reaction 2:
Pb2+(aq) + 2e- --> Pb(s) (E° = -0.13 V)
Notice that the second half-reaction needs to be reversed because we are using it as an oxidation half-reaction in the overall reaction.
Oxidation half-reaction (reversed reduction half-reaction 2):
Pb(s) --> Pb2+(aq) + 2e- (E° = +0.13 V)
Now, we can add the two half-reactions together to obtain the overall balanced reaction:
2Cr(s) + 3Pb2+(aq) --> 3Pb(s) + 2Cr3+(aq)
To determine the standard cell potential for the overall reaction, we add the reduction potentials of the half-reactions:
E°cell = E°cathode - E°anode
E°cell = (+0.13 V) - (-0.74 V)
E°cell = 0.13 V + 0.74 V
E°cell = 0.87 V
Therefore, the standard cell potential for the given reaction is 0.87 V.