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ONE half cell in a voltaic cell is constructed from a silver wire dipped in a silver nitrate solution of unknown concentration. its other half cell consists of a zinc electrode dipped in 1.0 M solutionof zinc nitrate. a voltage of 1.48 is measured for this cell. calculate concentration of silver nitrate. {standard electrode potentials are given: Zn2+\\\\Zn= -0.76 V . Ag+\\\\Ag = 0.80 V.}

Himanshu Gupta , 14 Years ago
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anser 2 Answers
SAGAR SINGH - IIT DELHI

Last Activity: 14 Years ago

Dear student,

eq 8

where, for a generalized equation of the form:

eq 9

The capital letters A, B, M and N in equation represent respectively the reactants and products of a given reaction while the small letters represent the coefficients required to balance the reaction.

eq 10

At equilibrium, DG = 0 and Qreaction corresponds to the equilibrium constant (Keq) described earlier.

In the case of an electrochemical reaction, substitution of the relationships DG = -nFE and D

G0 = -nFE0 into the expression of a reaction free energy and division of both sides by -nF gives the Nernst equation for an electrode reaction:

eq 11

Combining constants at 25oC (298.15 K) gives the simpler form of the Nernst equation for an electrode reaction at this standard temperature:

eq 12

In this equation, the electrode potential (E) would be the actual potential difference across a cell containing this electrode as a half-cell and a standard hydrogen electrode as the other half-cell. Alternatively, the relationship in equation can be used to combine two Nernst equations corresponding to two half-cell reactions into the Nernst equation for a cell reaction:

eq 13

Muskan Rimjha

Last Activity: 3 Years ago

At cathode
Ag+ + e  -> Ag
At anode
Zn -> Zn2+ + 2e
The net reaction
Zn + 2Ag+ -> Zn2+ +2Ag
E0cell =ECathode - Eanode
           = 0.80+0.76
           = 1.56V
Zn(NO3)2 ->    Zn2+       +          2NO3-
  1.0mol         1.0mol               2.0 mol
ECell=E0cell -( 0.0591/n) (Log [Zn2+]/[Ag+]2)
 
1.48=1.56-(0.0591/2)(Log[1.0]/[Ag+]2)
1.48-1.56=-0.0295(Log 1 - Log [Ag+]2)
-0.08= -0.0295(- Log[Ag+]2
Log[Ag+]2=-2.712
[Ag+]2= Antilog(-2.712)
[Ag+]2=5.152×10-2
[Ag+] = 0.225 M
      AgNO3  ->      Ag+      +      NO3-
Then the concentration of silver nitrate is 0.225 M
 
 
                       

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