3-12
9 Questions
Given: E (Cu Cu-)0.34V A zinc electrode is placed in a 0.1 M solution at 25°C. Assuming that
the salt is 20% dissociated at this dilutions calculate the electrode potential. Eo (Zn2 |Zn) --0.76V
Q.3 Q4 From the standard potentials shown in the following diagram, calculate the potentials Ei
and E2 Q.5 Calculate the EMF of the following cell ZnIZn2 + (001 M)!! Zn2 + (0.1 M) Zn at
298 K. Q.6 Calculate the EMF of the cell, at 25°C, if the concentrations of the zinc amalgam
are: c1 10g per 100g of mercury and c2 lg per100 g of mercury Q.7 Calculate pH using the
following cell Pt (112) l H+ (x M) 1 1 H+ ( 1 M) I Pt (H2) if Ecell = 0.2364 V 1 atm 1 atm
Calculate the equilibrium constant for the reaction Fe Fe3+ + Ce3+, [given: Ece Icet 1.44V;E Fe
/Fe -0.68V] Q.8 Calculate the equilibrium constant for the reaction Fe + CuSO4 ? FeSO4 + Cu at
25°C. Given E° (Fe/Fe2t) 0.44V, E° (Cu/Cu)--0.337V. Q.9 The standard reduction potential
at 25°C for the reduction of water 2H20+ 2e- H2 + 20H-is-0.8277 volt. Calculate the
equilibrium constant for the reaction 21,0#H,O+ Q.10 + OH-at 25°C. At 25°C the value of K
for the equilibrium Fe3+ + Ag ? Fe2+ + Ag" is 0.531 mollitre. The standard electrode potential
for Ag+ + e-? Ag is 0.799V. What is the standard potential for Fe3+ + e Fe2+ ? Q.11 Q12 For
the reaction, 4A1(s) + 302(g) + 6H20+ 4 OH-# 4 [Al(OH)4-1: Ew" 2.73 V. If ?G; (OH' ) -1 57
kJ mol-l and &G; f (H2O)--237.2 kJ mol-1, determine ?Gr [Al (OH),1. VMC
Solution
Q:3. Sol:-
We know that Reduction half cell reaction of Zn is :
Zn 2+
(aq) + 2e -
-------------> Zn(s) , E 0
Zn
2+
/Zn = - 0.76 V
Now Nernst equation for this equation is :
E Zn
2+
/Zn = E 0
Zn
2+
/Zn - 0.0591 / n log 1 / [Zn 2+
] ........... (1)
E Zn
2+
/Zn = Non-Standard Reduction potential = ?
E 0
Zn
2+
/Zn = Standard Reduction potential = - 0.76 V (given)

n = number of electrons gain = 2
Now because salts is 20 % dissociated only , therefore concentration of Zn 2+
i.e [Zn 2+
] will be :
[Zn 2+
] = 20 % of 0.1 M
= 20 x 0.1 M / 100
= 0.02 M
Now on substituting all these values in equation (1) , we have
E Zn
2+
/Zn = -0.76 V - 0.0591 / 2 log 1 / 0.02 M
E Zn
2+
/Zn = - 0.76 V - 0.02955 log 50
E Zn
2+
/Zn = - 0.76 V - 0.02955 x 1.69897
E Zn
2+
/Zn = - 0.76 V - 0.05020
E Zn
2+
/Zn = - 0.81 V
Hence electrode potential i.e E Zn
2+
/Zn = - 0.81 V