Equation Balance: Kaplan wrong?

[phosphatase]

---

Members do not see this ad.

kaplan question:

1. Mn + H2O → MnO4– + H+
When the reaction above is balanced, how many
electrons must be added to the right side of the
equation?
A. 0 moles
B. 1 moles
C. 2 moles
D. 5 moles
E. 7 moles

My solution: 2Mn + 8H2O → (2MnO4–) + (2H+) + (7H2) So answer is A
Kaplan answer: E

Could anyone correct me or confirm me please? thanks

By the way, how could a BALANCED equation has nude electrons on one side? thanks, I am totally confused

 

[americanpierg]

kaplan question:

1. Mn + H2O → MnO4– + H+
When the reaction above is balanced, how many
electrons must be added to the right side of the
equation?
A. 0 moles
B. 1 moles
C. 2 moles
D. 5 moles
E. 7 moles

My solution: 2Mn + 8H2O → (2MnO4&#821😉 + (2H+) + (7H2) So answer is A
Kaplan answer: E

Could anyone correct me or confirm me please? thanks

By the way, how could a BALANCED equation has nude electrons on one side? thanks, I am totally confused

In the balanced equation, how many electrons must be added to the right side. look at the oxidation of Mn since the oxidation number of hydrogen doesnt change. no oxidizing agent so the electrons are "nude" as you call it

 

[wantVCUdental]

kaplan question:

1. Mn + H2O → MnO4– + H+
When the reaction above is balanced, how many
electrons must be added to the right side of the
equation?
A. 0 moles
B. 1 moles
C. 2 moles
D. 5 moles
E. 7 moles

My solution: 2Mn + 8H2O → (2MnO4–) + (2H+) + (7H2) So answer is A
Kaplan answer: E

Could anyone correct me or confirm me please? thanks

By the way, how could a BALANCED equation has nude electrons on one side? thanks, I am totally confused

Kaplan is correct, and you are not balancing nude electron, you can't just balance redox like regular equations, you must separate the different species.
This problem, i did it in the following way
Mn -> MnO4-
H2O->H+

Let's balance the first one,
Mn + 4H2O -> MnO4- + 8H+ +7e
Now the second one
H2O + H+ -> H2O + H+ (no electrons involved)

Now add the two together

Mn + 4H2O + H2O + H+ -> MnO4- + 8H+ +7e + H2O + H+
therefore we need 7 moles of electrons
Hope this helps

 

[phosphatase]

thanks for your replys!

To americanpierg: H+ is the oxidizing agent, and it's reduced into H2 in my method.

To wantVCUdental: don't you think (7H+) + (7e-) --> 3.5 H2 ?
To wantVCUdental: Have you ever seen a balanced full ReDox equation with (e-) on either side? If yes, could you please show me a link? I googled but got my confused.

Yes, I understand there is 7e transferd for per Mn atom. But in my mind, whatever the e number is, it shouldn't be shown on the final equation and must be balanced by H+, OH-, H2O.

Please correct me harshly.

 

[americanpierg]

thanks for your replys!

To americanpierg: H+ is the oxidizing agent, and it's reduced into H2 in my method.

You can't create an entirely new species (H2) just to create an oxidizing agent. You only add H+, OH-, and H20 becuase of the acidic/basic solution allows for it. In the original equation, there is no oxidizing agent, the electrons have to go somewhere, so you just write them in.

 

[phosphatase]

To americanpierg: I don't agree. How could H+ AND E- stay in the same solution peacefully? Don't you think H2 is formed? Or else what is reduced?

 

[americanpierg]

To americanpierg: I don't agree. How could H+ AND E- stay in the same solution peacefully? Don't you think H2 is formed? Or else what is reduced?

Nothing is reduced. This isn't a redox equation. From what is given in the equation, the electrons will most likely be allowed to escape from the solution, probably through a wire to produce a current. Maybe this equation doesn't even exist in the real world, and they just want to see if you understand the basics behind balancing charges.

You don't know that H2 is formed because the equation doesn't say H2 is formed. That's like wirting in Cl2 gas into an equation because you are given Cl- and no reducing agent.

 

[phosphatase]

thanks americanpierg, just hope this doesn't show up on DAT.

btw, could you please have a look at this?
http://forums.studentdoctor.net/showthread.php?t=673001