When combined with cupric ions (aq, 1M) which will produce the largest standard cell voltage? (Use data from table 20.1, and Appendix D page A28-A29 in Petrucci)
1. ) Mn(s)
2. ) MnCl2(aq)
3. ) KMnO4(aq)
4. ) MnO2(s)
the answer i got was MnO2,,,,but im not sure if its right...can someone confirm it?
thx
Hey how did you find the stand cell volutage of MnCl2 and KMNO4?
To determine which substance will produce the largest standard cell voltage when combined with cupric ions, we need to compare the reduction potentials (E°) of each substance.
Based on Table 20.1 and Appendix D in Petrucci, we can find the reduction potentials for each substance:
1. Mn(s) --> Mn2+(aq) + 2e- E° = -1.18 V
2. MnCl2(aq) --> Mn2+(aq) + 2e- E° = -1.05 V
3. KMnO4(aq) --> MnO4-(aq) + e- E° = 1.51 V
4. MnO2(s) --> Mn2+(aq) + 4H+(aq) + 2e- E° = 1.70 V
Comparing the reduction potentials, we see that MnO2(s) has the highest reduction potential (E° = 1.70 V). Therefore, when combined with cupric ions (Cu2+), MnO2(s) will produce the largest standard cell voltage.
Therefore, your answer of MnO2 is correct.
To determine which substance will produce the largest standard cell voltage when combined with cupric ions (Cu2+), we need to refer to the given information in Table 20.1 and Appendix D pages A28-A29 in Petrucci.
1. Mn(s): Referring to the table, we need to find the reduction potential for the half-reaction:
Mn2+(aq) + 2e- → Mn(s)
If we cannot find this specific reaction in the table, we need to look for the reduction potential for Mn2+:
Mn2+(aq) + 2e- → Mn(s)
From Appendix D (page A28-A29), we find the reduction potential of Mn2+ to be -1.18 V.
2. MnCl2(aq): Since the reduction potential for Mn2+ is already given and it is the same species in MnCl2, the reduction potential will remain the same as -1.18 V.
3. KMnO4(aq): We need to find the reduction potential for the half-reaction:
MnO4-(aq) + 4H+(aq) + 3e- → MnO2(s) + 2H2O(l)
From Table 20.1, we find the reduction potential of this reaction to be +1.51 V.
4. MnO2(s): Since the given option is a solid, we need to convert it into an aqueous form and find the reduction potential for the half-reaction:
MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)
From Table 20.1, we find the reduction potential of this reaction to be +0.56 V.
Comparing the reduction potentials, we can see that KMnO4(aq) has the highest reduction potential of +1.51 V, followed by MnO2(s) with +0.56 V.
Therefore, the correct answer is 3.) KMnO4(aq) as it will produce the largest standard cell voltage when combined with cupric ions (Cu2+).