Wednesday 15 May 2013

Write the Nernst equation and emf of the following cells at 298 K



Q.5:Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)
(ii) Fe(s) | Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s)
(iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s)
(iv) Pt(s) | Br2(l) | Br−(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).
Solution:
(i)Write the Nernst equation and emf of the following cells at 298 K
Net reaction 
Mg(s) +Cu+2(aq)↔ Mg+2(aq) + Cu(s)
Nernst equation
There are two electron are transferring so that n = 2
Eocell = Eo right  – Eoleft
Eocell =+0.34 – (–2.37)V
 Eocell =+2.71V
Plug the value we get
Concentration of solid substance is 1 always so that [Mg] = [Cu] =1

(ii)  Write the Nernst equation and emf of the following cells at 298 K
Net reaction 
Fe(s) +2H+(aq)↔ Fe+2(aq) + H2(g)
Nernst equation
There are two electron are transferring so that n = 2
Eocell = Eo right  – Eoleft
Eocell =0– (–0.44)V
 Eocell =+0.44 V
Plug the value we get
Concentration of solid substance is 1 always so that [Fe] = [H2] =1

(iii) Write the Nernst equation and emf of the following cells at 298 K
 Net reaction 
Sn(s) +2H+(aq)↔ Sn+2(aq) + H2(g)
Nernst equation
There are two electron are transferring so that n = 2
Eocell = Eo right  – Eoleft
Eocell =0– (–0.14)V
 Eocell =+0.14 V
Plug the value we get
Concentration of solid substance is 1 always so that [Sn] = [H2] =1
(iv) ) Write the Nernst equation and emf of the following cells at 298 K
  Net reaction 
2Br(aq) +2H+(aq)↔ Br2(l)+ H2(g)
Nernst equation
There are two electron are transferring so that n = 2
Eocell = Eo right  – Eoleft
Eocell =0– (1.08)V
 Eocell –1.08V
Plug the value we get
Concentration of solid substance is 1 always so that [Br2] = [H2] =1

12 comments:

  1. can any one teach me that in above question why 1/([Br]2[H]2
    in question no. iv

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  2. molar concentration of Br2(l) = H2(g) = 1
    and use product / reactant there

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  3. Thanks for your reply .
    Sir my question was something different , May be I have not explained properly . I am explaining it again
    In Other equations e.g. In Equation no. (iii) Log Kc value was [Sn2+]/ [H+]2 (i.e. "Sn " is numerator while "H " is denominator )
    While in Equation no. (iv) Log Kc value was 1/ [Br-]2[H+]2 ( i.e. Bothe are denominator.
    Sir I wanted to know the reason behind this
    Please tell in detail, I will be very grateful to you .

    ReplyDelete
    Replies
    1. I also had same question - because we take products of reaction in numerator and reactants in denominator .

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    2. Because both are undergoing reduction so both concentration will be in denominator

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    3. It means the both reactions are on cathode so E°cell=Ecathode-Eanode
      Then E°cell comes positive not negative???

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  4. Because both are undergoing reduction so both concentration will be in denominator

    ReplyDelete
  5. how are we getting the values of E not cell

    ReplyDelete
    Replies
    1. Check table 3.1 in 12th ncert textbook, you will get the standard electrode potential of different species. In the equation "E0cell= E0cathode - E0anode" substitute the values from the table.

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