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 MnCl2,,,,but im not sure if its right...can someone confirm it?
what is going to reduce Cu2+ ions then if all you have are Cl- ions (does not like being oxidized) and Mn2+ ions (wants to be reduced not oxidized)
so would the answer be Mn (s)?
To determine which substance will produce the largest standard cell voltage when combined with cupric ions (Cu2+(aq), 1M), we need to compare the reduction potentials of these substances.
According to table 20.1 in Petrucci, the standard reduction potentials (E°) of the substances involved are as follows:
1. Mn(s) = -1.18V
2. MnCl2(aq) = -1.22V
3. KMnO4(aq) = 1.51V
4. MnO2(s) = 1.23V
By comparing these values, we can see that KMnO4(aq) and MnO2(s) have the highest reduction potentials. However, since we are combining them with cupric ions (Cu2+(aq)), we need to consider the overall reaction taking place.
The half-reactions for the reduction of cupric ions (Cu2+) and the reduction of each substance are as follows:
Cu2+(aq) + 2e- → Cu(s)
Mn(s) → Mn2+(aq) + 2e-
MnCl2(aq) + 2e- → Mn(s) + 2Cl-(aq)
KMnO4(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)
From the half-reactions, we can deduce that Mn(s) and MnO2(s) are both reduced, while KMnO4(aq) is oxidized, and MnCl2(aq) doesn't have any net change.
Since standard cell voltage is determined by the difference in reduction potentials of the two half-reactions, the largest standard cell voltage will occur when the reduction of MnO2(s) is paired with the reduction of Cu2+(aq).
Therefore, the correct answer is 4) MnO2(s).
To determine which compound will produce the largest standard cell voltage when combined with cupric ions, we can use the standard reduction potentials found in Appendix D of Petrucci (page A28-A29). The standard reduction potentials for the compounds listed in the question are as follows:
1. Mn(s): -1.18 V
2. MnCl2(aq): -1.16 V
3. KMnO4(aq): 0.56 V
4. MnO2(s): 1.23 V
The standard cell voltage can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode. In this case, the anode is the copper half-reaction (Cu2+ + 2e- -> Cu) with a standard reduction potential of +0.34 V (taken from table 20.1), and the cathode is the reduction of Mn or its compounds.
For each compound, we subtract the reduction potential of the copper half-reaction from the reduction potential of the compound to determine the overall standard cell voltage:
1. Mn(s): -1.18 V - (+0.34 V) = -1.52 V
2. MnCl2(aq): -1.16 V - (+0.34 V) = -1.50 V
3. KMnO4(aq): 0.56 V - (+0.34 V) = 0.22 V
4. MnO2(s): 1.23 V - (+0.34 V) = 0.89 V
Based on the calculations, the compound that will produce the largest standard cell voltage when combined with cupric ions is MnO2(s) with a standard cell voltage of 0.89 V. Therefore, the answer is option 4: MnO2(s).
Please note that these calculations are based on standard conditions and may not accurately represent actual conditions in a real system.