Calculate the standard cell potential for each of the electrochemical cells?
What would the answers be to something like this? If someone could work out the answer I would really appreciate it. My online hw keeps telling me its wrong.
2Ag+(aq)+Pb(s)→2Ag(s)+Pb2+(aq)
E∘cell......V?
2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s)
E∘cell......V?
O2(g)+4H+(aq)+2Zn(s)→2H2O(l)+2Zn2+(aq)
E∘cell......V?
John, did you read the post I left for you? If you will show your work we can spot the problem. Otherwise we're working in the dark.
To calculate the standard cell potential (E°cell) for each electrochemical cell, you need to use the standard reduction potentials (E°red) of the half-reactions involved in each cell. The standard reduction potential is a measure of the willingness of a species to be reduced (gain electrons) compared to the standard hydrogen electrode (assigned as 0.00 V).
Let's work through the calculations for each of the given electrochemical cells:
1. 2Ag+(aq) + Pb(s) → 2Ag(s) + Pb2+(aq)
First, identify the half-reactions involved in the cell:
- Reduction half-reaction: Ag+(aq) + e- → Ag(s) (with E°red = +0.80 V)
- Oxidation half-reaction: Pb(s) → Pb2+(aq) + 2e- (with E°red = -0.13 V)
The standard cell potential (E°cell) is the difference between the reduction potential of the cathode and the reduction potential of the anode. Therefore,
E°cell = E°red(cathode) - E°red(anode)
E°cell = (+0.80 V) - (-0.13 V) = +0.93 V
So, the standard cell potential for this electrochemical cell is +0.93 V.
2. 2ClO2(g) + 2I-(aq) → 2ClO-(aq) + I2(s)
Again, identify the half-reactions:
- Reduction half-reaction: ClO2(g) + 2e- → ClO-(aq) (with E°red = +1.08 V)
- Oxidation half-reaction: 2I-(aq) → I2(s) + 2e- (with E°red = +0.54 V)
E°cell = E°red(cathode) - E°red(anode)
E°cell = (+1.08 V) - (+0.54 V) = +0.54 V
So, the standard cell potential for this electrochemical cell is +0.54 V.
3. O2(g) + 4H+(aq) + 2Zn(s) → 2H2O(l) + 2Zn2+(aq)
Let's find the half-reactions:
- Reduction half-reaction: O2(g) + 4H+(aq) + 4e- → 2H2O(l) (with E°red = +1.23 V)
- Oxidation half-reaction: 2Zn(s) → 2Zn2+(aq) + 4e- (with E°red = -0.76 V)
E°cell = E°red(cathode) - E°red(anode)
E°cell = (+1.23 V) - (-0.76 V) = +1.99 V
So, the standard cell potential for this electrochemical cell is +1.99 V.
Make sure to check the signs (both of the reduction potentials and their difference) as they determine the direction of the cell reaction. If you've inputted the values correctly and still receive an error, it's possible that the online homework system may have different conventions or significant figures requirements, so check the problem statement carefully or consult your instructor if needed.