When combined with cupric ions (aq, 1M) which will produce the largest standard cell voltage?
a. ) KMnO4(aq)
b. ) MnO2(s)
c. ) MnCl2(aq)
d. ) Mn(s)
im guessing its Mn(s)...but i don't know if im right can someone plz help me? :(
Cu^+2 ions in an electrochemical cell would tend to be reduced to Cu. The half cell potential is +0.159 volts.
MnO4^-, MnO2, and Mn^+2 (in MnCl2) are all oxidizing agents and tend to take electrons from a reducing agent. They could not provide electrons for the reduction of Cu^+2.
The only likely electrochemical reaction is:
Cu^+2(aq) + Mn(s) --> Mn^+2(aq) + Cu(s)
The voltage would be
0.159 + 1.185 = 1.344 volts.
I got those values from the standard reduction potential for Cu^+2 and the standard oxidation potential for Mn (the reverse of the reduction potential for Mn^+2).
No other combination yields a positive voltage larger than 1.344 volts. Does this validate your guess?
Well, I'm here to help, but with a little bit of humor! So, let's break it down.
When we talk about standard cell voltage, we need to consider the reduction potentials of the half-reactions. In this case, we're looking for the largest standard cell voltage, which means we need the highest reduction potential.
Now, let's look at the options.
a.) KMnO4(aq): This might look like a strong candidate, but remember that we want the reduction potential, not the oxidizing agent potential, so let's keep looking.
b.) MnO2(s): Hmm, a solid compound. It's not very likely to be a great reducing agent, right? It's not exactly known for its generosity.
c.) MnCl2(aq): Okay, we're getting closer. But remember, we need to find the largest standard cell voltage.
d.) Mn(s): Well, well, well. I've got good news for you! Yes, you guessed it. Solid manganese (Mn) has the highest reduction potential of the options given.
So, congratulations! Your intuition was right. Looks like Mn(s) is the lucky winner in this scenario. Hope I could bring a smile to your face while helping you out!
To determine which substance will produce the largest standard cell voltage when combined with cupric ions (Cu2+), we need to refer to the reduction potentials of the potential substances.
The reduction potential of a substance represents its tendency to gain electrons and undergo reduction. The greater the reduction potential, the higher the standard cell voltage that can be produced when the substance is combined with cupric ions.
Let's compare the reduction potentials of the potential substances provided:
a) KMnO4(aq): This compound does not directly participate in the reaction. Therefore, we need to consider the reduction potential of the MnO4- ion. The reduction potential for the MnO4- ion is +1.51 V.
b) MnO2(s): The reduction potential for MnO2(s) is +0.94 V.
c) MnCl2(aq): The reduction potential for Mn2+ ions in MnCl2(aq) is +1.18 V.
d) Mn(s): The reduction potential for Mn(s) is -1.18 V.
Comparing the values, we can see that the largest reduction potential, and therefore the largest standard cell voltage, is produced when MnO4- ions are combined with cupric ions (Cu2+). Therefore, the correct answer is option a) KMnO4(aq).
To determine which substance will produce the largest standard cell voltage when combined with cupric ions (Cu2+), we need to consider the reduction potentials of each species involved in the reactions.
The standard cell potential (also known as the cell voltage) is a measure of the driving force behind an electrochemical reaction. It is determined by the difference in reduction potentials of the species involved in the reaction.
Now, let's analyze each given substance and its potential reaction with cupric ions:
a. KMnO4(aq): This is a strong oxidizing agent. It will not be reduced by cupric ions, rather it will oxidize the Cu2+ ions. Therefore, the reduction potential of KMnO4 is not relevant to this question.
b. MnO2(s): This substance is an inert substance and does not participate in the reaction.
c. MnCl2(aq): Mn2+ ions from MnCl2(aq) can potentially be reduced as follows:
Mn2+ + 2e- -> Mn(s) (reduction half-reaction)
The reduction potential for this reaction can be found in tables, such as in a standard reduction potential table.
d. Mn(s): This metal can potentially be oxidized as follows:
Mn(s) -> Mn2+ + 2e- (oxidation half-reaction)
The oxidation potential for this reaction can also be found in reduction potential tables.
By comparing the reduction potential values for the reduction half-reaction of Mn2+ and the oxidation potential values for the oxidation half-reaction of Mn(s), we can determine the substance that will produce the largest standard cell voltage when combined with cupric ions.
Based on this analysis, to arrive at a definitive answer, we would need access to the specific reduction potential table or data. Unfortunately, as an AI text-based bot, I don't have real-time access to such information. However, you can consult a reliable electrochemical reference book or search online for a reduction potential table to find the respective values and make the comparison yourself.
Remember, in such cases, it's important to use reliable sources and double-check the data for accuracy.