Calculate the standard free energy change for the following reaction at 25°C.
Au(s) + Ca2+(aq.1M) → Au3+(aq. 1M) + Ca(s)
Au2+/Au = – 0.403 V.
Au2+/Cu = – 0.763 V.
Predict whether the reaction will be spontaneous or not at 25°C. Which of the above two half cells will act as an oxidizing agent and which one will be a reducing agent?

E° = E°Ca2+/Ca - E°Au3+/Au      = -287 - (1.50)      = -2.87 - 1.50 = 4.37 VrG° = nFE°                     

            = 6×96500×(-4.37) = 2530.23 kg
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Calculate the equilibrium constant for the reaction:
Cd2+(aq) + Zn(s) → Zn2+(aq) + Cd(s)
if E°Cd2+/Cd = – 0.403 V.
zn2+/zn = – 0.763 V


E0cell = E0(Cd2+/Cd) - E0(Zn2+/Zn)           = -0.403 - (0.763) = 0.36 VE0cell = 0.059nlog Klog K = n+E0cell0.059log K  = 2+0.36.059 = 12.2or     K = 1.585 × 1012.
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Tarnished silver contains Ag2S. Can this tarnish removed by placing silver in an aluminium pan containing an inert electrolytic solution such as NaCl. The standard electrode potential for half reaction.

Tarnished silver contains Ag2S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as NaCl. Explain the electrochemical principle for this procedure. (The standard reduction potential for the half-cell reaction Ag2S(s) + 2e- ---> 2Ag(s) + S2-(aq) is -0.71V.)
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When a current of 0.75 A is passed through a CuSO4 solution for 25 min, .36 g of copper is deposited at the cathode. Calculate the atomic mass of copper.

Time  = 25 min  = 25 x 60 sec
Current     = 0.75 A
Cu2++2e-      Cu
Electricity passed  = 25 x 60 x 0.75  = 1125 C of electricity deposit copper  = 0.369 g. 2 x 96500 C
of electricity will deposit copper.
M= ZxIxT

= 2 × 96500 × 0.369 1125
Atomic mass of copper = 63.3 u.

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The conductivity of 0.001 M acetic acid is 4 x 10–5 s/cm. Calculate the dissociation constant of acetic acid, if λ°m for acetic acid is 390.5 s cm2 mol. 

concentration =0.001M
conductivity =4 x10-5 s/cm
 = k×1000Mk = 4×10-5s/cmM = 0.001 M = 4×10-5×10000.001      = 4×10-5×103×103      = 40 S cm2 mol-1.0 = 390.5 S cm2/molα = 0 =40390.5 = 0.102K = 2    = 0.001 × (0.102)2     = 1.04 × 10-5.
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