Question

Q7
a) Calculate the EMF of the following cell :
Zn/Z2+(1M)//CD2/Cd (1M) [Given E' of Zn/Zn = 0.76V and Eº of Cd/ca+2=0.40 VI
b) Given the standard electrode potentials
K+/K=-2.93V, Agt/Ag=0.80V, Hg2+/Hg= 0.79, Mg2/Mg= -2.37V,
Cr+3/Cr= -0.74V. Arrange these metals in increasing order of their reducing power.​

Answer 1

Answer:

Explanation:

emf of cell= reduction potential of (cathode- anode)

                 = -0.40-(-0.76)

                 = 0.26V

reducing power ∝ oxidiation potential ∝ $\frac{1}{Reduction Potential}$

∴ order is : K>Mg>Hg>Ag

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