Question

Check an Answer Please ? - See Reply

Answer 1

\[MnO^- + 8H^+ 5e^1 \rightarrow Mn ^{2+} + 4H _{2} O\]
\[NO _{3}^ {-} + 3H ^{+} 2e ^{-} \rightarrow HNO _{2} + H _{2}O\]
Balancing them of reaction MnO4- with HNO2
Then calculating Gibbs Free Energy

Answer 2

the 5e should be a negative sine btw, the E/V for MNO4 is +1.51 and for HNO2 is +0.94

Answer 3

\[2MnO4^{-} + 16H ^{+} \rightarrow 5NO _{3} ^{-} + 15 H ^{+}\]
and n for me was 10

Answer 4

So good so far... which reactions did you donate I and II in order to find right and left?

Answer 5

MnO4 should also be on first equation , sorry first day on this site :/

Answer 6

Then -10 * 96485 * (Not Sure as to the Electrode as cant tell oxidation/reduction) Attempting the I and II now

Answer 7

+0.94 - 1.51 Im presuming is the order or II - I ?

Answer 8

of*

Answer 9

Emmm...
I want it to be +1.51 - (+0.94)
I: 2 MnO4- + 16 H+ + 10 e- ->2 Mn(2+) + 8 H2O
II: 5 NO3- + 15 H+ + 10 e- -> 5 HOH2 + 5 H2O

I - II:
The reaction for (-II):
5 HOH2 + 5 H2O ->5 NO3- + 15 H+ + 10 e-

We then get:
2 MnO4- + 16 H+ + 10 e- + 5 HOH2 + 5 H2O -> 5 NO3- + 15 H+ + 10 e- + 2 Mn(2+) + 8 H2O
Remove the things that goes out with each other:
2 MnO4- + 1 H+ + 5 HOH2 + -> 5 NO3- + 2 Mn(2+) + 3 H2O (redox)
So it most be:
if you are going to calculate Delta G for reaction above.
\[\LARGE E _{cell}=E(I)-E(II)\]

Answer 10

I: 2 MnO4- + 16 H+ + 10 e- ->2 Mn(2+) + 8 H2O E=+1.51
II: 5 NO3- + 15 H+ + 10 e- -> 5 NHO2 + 5 H2O E=+0.94
Please check this is the right information; crucial for the result.