Cu2+ + 2e → Cu E° = + 0.34 V
Ag+ + 1e → Ag E° = + 0.80 V
(i) Construct a galvanic cell using the above data.
(ii) For what concentration of Ag+ ions will the emf of the cell be zero at 25°C, if the concentration of Cu2+ is 0.01 M ? [log 3.919 = 0.593]


(i) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)
(ii) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)

(i) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)(ii) Cu(s) | Cu2+ (aq) ||


(i) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)(ii) Cu(s) | Cu2+ (aq) ||

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The measured resistance of a conductance cell containing 7.5 x 10–3 M solution of KCl at 25° was 1005 ohms. Calculate (a) specific conductance (b) Molar conductance of the solution. Cell constant = 1.25 cm–1.


M = 7.5 × 10-3R = 1005 ohms.
(a) Specific conductance (K)
             = 1R×1a where 1a is cell constant.
K=11005 Ohm×1.25 cm-1K = 1.2437×10-3ohm-1cm-1    = 1.2437×10-3S cm-1

(b) 
        Λm= 1000×KM=1000×1.24377.5×10-3m     = 1.2437×10007.5=1.24377.5      = 165.826 Ω-1 cm2 mol-1Λm = 165.826 S cm2 mol-1.
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Calculate emf of the following cell at 298 K
Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)
(Given : E°Cu2/Cu = + 0.34 V; E° Mg2+/Mg = – 2.37 V)

Mg left parenthesis straight s right parenthesis space rightwards arrow space Mg to the power of 2 plus end exponent space plus space 2 straight e to the power of minus
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Silver is electrodeposited on a metallic vessel of total surface area 900 cm2 by passing a current of 0.5 amp for two hours. Calculate the thickness of silver deposited. [Given : Density of silver = 10.5 g cm–3, Atomic mass of silver = 108 amu, F = 96,500 C mol–1]


m = Z x I x t
or          m= 10896500×0.5×2×60×60

or                 m= 108×5965×10×2×6×6 = 4.03 g     m = V × d   4.03 = V×10.6 g cm-3
or                      V = 4.0310.6  cm3 = 0.39 cm3V = area × thickness
or      0.39 cm3 = 900 cm2 × thickness
Thickness = 0.39 cm3900 cm2 = 4.33 × 10-4 cm

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A 0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell is 0.367 cm–1, find out the specific and molar conductance of the sodium hydroxide solution.

(i) Resistance (k) = 3.16 Ω
Conductance (C) = 1R
         = 13.16 ohm = 0.0316 ohm-1
Specific conductance (k)
                                 = Conductance x cell constant
                                = 0.0316 ohm-1×0.367 cm-1= 0.0116 ohm- cm-1
(ii) Molar conductance (C)
                         = 0.05 M = 0.05 mol L-1= 0.05 mol1 L = 0.05 mol103 cm3= 0.05 × 10-3 mol cm-3
Molar conductance (Λm)
                          = kC=(0.0116 ohm-1 cm-1)(0.05×10-3 mol cm-3)= 232 ohm-1 cm2 mol-1.


                                 

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