Question

The reduction potential of hydrogen half cell will be negative if:

Answer 1

if p(H2) =a and [H+] = b then options are:
a=1atm,b=1M
a=2atm b=1M
a=2atm b=2M
a=1atm b=2M

Answer 2

@Rushwr @Cuanchi

Answer 3

do you have any idea?
are you familiar with the Nernst equation?

Answer 4

look at SAMPLE EXERCISE 20.13
in this page
http://wps.prenhall.com/wps/media/objects/3313/3392587/blb2006.html

Answer 5

@Cuanchi yes, i am familiar with Nernst eq. but i have a doubt what is the anode here?

Answer 6

@samigupta8 @imqwerty

Answer 7

@priyar it's just a half cell here so no anode and cathode is platinum electrode in which hydrogen is bubbled at 1 atm and the solution is 1M H+ concentration

Answer 8

What i m worried about is how will we get a negative potential of hydrogen through the value of b that u hve provided here ..it can be negative only if [H+] conc was lying in between
0 to 1(greater than 0 and less than 1)

Answer 9

You have to consider also the Pressure and calculate the Q. when the Q is smaller than 0 the E will be positive because the loq Q will be negative and if the Q > 0 , the log Q is positive and the E of the cell will be negative.
the anode is negative and is where the oxidation occurs
In this problem the anode is the standard H electrode and the cathode will be the non-standard one

Answer 10

Really @cuanchi

Answer 11

Won't it be like the way i suggested

Answer 12

@ganeshie8 @photon336

Answer 13

@ParthKohli

Answer 14

@Cuanchi please help

Answer 15

The reaction will be:
H+ + e- --> 1/2 H2
Eo = Eright - Eleft
= x -0 =x
and Q= p(H2)^1/2/[H+]
what will be Ecell (that is the RHS of Nernst equation)?

Answer 16

@Vincent-Lyon.Fr

Answer 17

@Kainui

Answer 18

a=2atm b=1M

Answer 19

I agree with @cuanchi...

Answer 20

@Cuanchi can you explain how you solved?

Answer 21

\[Q= \frac{ ([H+] electrode1)^{2} \times P _{H2} electrode2 }{([H+] electrode2)^{2} \times P _{H2} electrode1 }= \frac{ 1\times P _{H2} electrode2}{ [H+] electrode2)^{2} \times 1 }\]

we agree that the electrode 1 is the standard electrode

\[Q=\frac{ P _{H2} }{ [H+]_{2}^{2} }\]

when the Q is smaller than 0 the E will be positive because the loq Q will be negative and if the Q > 0 , the log Q is positive and the E of the cell will be negative.

Answer 22

what about Eo of cell (standard potential ) ? Nernst Eq has both Ecell and Eo cell.. can you tell me what will the value of both in this eq? coz thats where my doubt lies..

Answer 23

@Cuanchi

Answer 24

Eo is equal to "0" if the two half cells are the same composition (in standard conditions the concentration of the cells has to be 1M and the pressure 1atm by definition of standard conditions).

When you have a concentration cells, that means the only difference between the two cells is the concentration of the solutions or the pressure of the gases, the standard potential "Eo = 0" then you need the Nernst equation to calculate the E that it is dependent of the Q (Q=1 => log Q =0).
When two different half cells are in standard condition (1M, 1atm) the E = Eo and it is independent of the Q .

If you have both, different half cells composition and they are not at standard conditions you use the Nernst eq and you calculate the Eo with the standard potential of the half cells and the Q with the concentrations of products and reactants.

Answer 25

so these are concentration cells? so we use nernst eq with Eo=0 right? Thanks!

Answer 26

yes!!! definitely they are!