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?
See your other post above.
To calculate the standard cell potential (E°cell) for each of the electrochemical cells, you need to use the standard reduction potentials (E°red) of the half-reactions involved. The formula for calculating E°cell is:
E°cell = E°cathode - E°anode
Let's calculate the standard cell potential for each electrochemical cell:
1. 2Ag+(aq) + Pb(s) → 2Ag(s) + Pb2+(aq)
To find the values for Ag+ and Pb2+, we refer to the standard reduction potentials table. The reduction half-reactions involved are:
Ag+(aq) + e- → Ag(s) (E°red = +0.80 V)
Pb2+(aq) + 2e- → Pb(s) (E°red = -0.13 V)
Now, we plug these values into the formula:
E°cell = E°cathode - E°anode
E°cell = 0.00 V (since we have Ag(s) on both sides)
2. 2ClO2(g) + 2I-(aq) → 2ClO-(aq) + I2(s)
The reduction half-reactions involved are:
ClO2(g) + 2e- → ClO-(aq) (E°red = +0.96 V)
I2(s) + 2e- → 2I-(aq) (E°red = +0.54 V)
Now, we plug these values into the formula:
E°cell = E°cathode - E°anode
E°cell = 0.42 V
3. O2(g) + 4H+(aq) + 2Zn(s) → 2H2O(l) + 2Zn2+(aq)
The reduction half-reactions involved are:
O2(g) + 4H+(aq) + 4e- → 2H2O(l) (E°red = +1.23 V)
Zn(s) → Zn2+(aq) + 2e- (E°red = -0.76 V)
Now, we plug these values into the formula:
E°cell = E°cathode - E°anode
E°cell = 1.99 V
Please check if the values you inputted match the table values correctly. Double-checking is important to ensure accuracy.
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 and apply the Nernst equation.
Let's solve each case step by step:
1. 2Ag+(aq) + Pb(s) → 2Ag(s) + Pb2+(aq)
To find the E°cell for this cell, you need to look up the standard reduction potentials for each half-reaction involved.
The half-reactions are:
Ag+ + e- → Ag (reduction)
Pb2+ + 2e- → Pb (oxidation)
The standard reduction potentials for these half-reactions are:
E°red(Ag+/Ag) = +0.80 V
E°red(Pb2+/Pb) = -0.13 V
To get the overall cell potential, you subtract the reduction potential of the oxidation half-reaction from the reduction half-reaction:
E°cell = E°red(reduction) - E°red(oxidation)
E°cell = (0.80 V) - (-0.13 V)
E°cell = 0.93 V
Therefore, the standard cell potential for this electrochemical cell is 0.93 V.
2. 2ClO2(g) + 2I^-(aq) → 2ClO^-(aq) + I2(s)
For this cell, you need the standard reduction potentials for each half-reaction:
ClO2 + e- → ClO^-(reduction)
I2 + 2e- → 2I^-(oxidation)
The standard reduction potentials for these half-reactions are:
E°red(ClO2/ClO^-) = +1.28 V
E°red(I2/I^-) = +0.54 V
Again, subtract the reduction potential of the oxidation half-reaction from the reduction half-reaction:
E°cell = E°red(reduction) - E°red(oxidation)
E°cell = (+1.28 V) - (+0.54 V)
E°cell = 0.74 V
Therefore, the standard cell potential for this electrochemical cell is 0.74 V.
3. O2(g) + 4H+(aq) + 2Zn(s) → 2H2O(l) + 2Zn2+(aq)
The half-reactions involved here are:
O2 + 4H+ + 4e- → 2H2O (reduction)
Zn → Zn2+ + 2e- (oxidation)
The standard reduction potentials for these half-reactions are:
E°red(O2/H2O) = +1.23 V
E°red(Zn2+/Zn) = -0.76 V
Now, subtract the reduction potential of the oxidation half-reaction from the reduction half-reaction:
E°cell = E°red(reduction) - E°red(oxidation)
E°cell = (+1.23 V) - (-0.76 V)
E°cell = 1.99 V
Therefore, the standard cell potential for this electrochemical cell is 1.99 V.
Make sure to double-check the reduction potentials you're using and the sign conventions. If your online homework still marks your answers as incorrect, it's possible that there might be a small rounding error or a different convention being used. In such cases, reach out to your instructor for further guidance.