Calculate the cell emf at 25° C for the following cell:
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
[Given E°Ni2+/Ni = – 0 25 V, E° Cu = + 0.34 V, 1 F = 96500]
Calculate the maximum work that can be accomplished by operation of this cell.


Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
At anode Ni(s) → Ni2+ (aq) + 2e
At cathode Cu2+ + 2e → Cu(s)
Net cell reaction

Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)At anode Ni(s) → Ni

Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)At anode Ni(s) → Ni

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(a) Calculate the electrode potential of silver electrode dipped in 0.1 M solution of silver nitrate at 298 K assumimg AgNO3 to be completely dissociated. The standard electrode potential of Ag+/Ag is 0.80 V at 298 K.
(b) At what concentration of silver ions will this electrode have a potential of 8.0 V?

(a) Ag++e- Ag(s)
     EAg+/Ag = E0Ag+/Ag - 0.05911log1Ag+               = 0.80 - 0.05911log10.1               = 0.80-0.0591 log 10               = 0.80-0.0591×1 = 0.7409 V
(b) Now,   
                EAg+/Ag = 0or  0.80-0.05911log1Ag+=0or                           log[Ag+] = 0.800.0591or                                   Ag+ = 3.438 × 1013M.

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Calculate the cell e.m.f. at 25° C for the following cell:
Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn (s)
[Given E°Mg2+/Mg = – 2.34 V, E°Sn2+/Sn = – 0.136V, 1 F = 96,500 C mol–1]
Calculate the maximum work that can be accomplished by the operation of this cell.

Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn(s)
Mg(s) → Mg2+ (aq) + 2e
(oxidation at anode)
Sn2+(aq) + 2e → Sn(s)
(reduction at cathode)

Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn(s)Mg(s) → Mg2+ (aq) +

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Calculate the e.m.f. of the cell in which the reaction is
Mg(s) + 2Ag+(aq)  Mg2+(aq) + 2Ag(s)When    Mg2+ = 0.130 Mand           Ag+ = 1.0 × 10-4M.    [Given EMg2+/Mg = -2.37 V and E°Ag+/Ag = 0.80 V]


Mg left parenthesis straight s right parenthesis space rightwards arrow space space space Mg to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 straight e to the power of minus



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The emf of a cell corresponding to the reaction
Zn(s) + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) (1 atm) is 0.28 V at 15° C.Write the half cell reactions and calculate the pH of the solution at the hydrogen electrode.


We have given the cell reaction
Zn(s) + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) (1 atm) 
thus 
E0Zn2+/Zn = -0.76 V,   E0H+/H2 = 0
Half call reaction will be
            Zn2++2e-  Zn                    ...(i)
     H++e- = 12 H2
 or                2H+ + 2e-  H2         ...(ii)
        EZn/Zn2+ = E°Zn/Zn2+-RTnFlnZn2+Zn
Here          
          R = 8.314 J mol-1 log-1,T = 298 K,F = 96500 coulomb n = 2,E0Zn/Zn2+ = 0.76.
Therefore,
EZn/Zn2+ = 0.76 - 8.314 × 2982 × 96500 ln 0.11                   = 0.79 V
Similarly,
          EH+/H2 = E0H+/H2 - RTnFlnH2H+2             =0-0.314×2982×96500 ln IH+2             = 0.05915 log10H+              = 0.05915 pH                                       [ - log10H+ = pH]
Now since
            E = EZn/Zn2+ + EH+/H2    = 0.28 = 0.79 - 0.05915 pHpH = 0.150.05915 = 8.62.

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