Single Choice

To the Daniel cell, if $$ZnSO_{4}$$ is added to the left hand side electrode, then cell emf:

Aincreases
Bdecreases
Correct Answer
Cdoes not change
Dfirst increases & then decreses

Solution

The reduction potential of anode is directly proportional to the concentration of the electrolyte. When zinc sulphate is added to the solution present in the anode compartment, then the concentration of the solution increases. This increases the reduction potential of anode. Hence, the emf of the cell decreases as $$ { E }_{ cell } = { E }_{ cathode } - { E }_{ anode } $$. Hence, option $$B$$ is correct.


SIMILAR QUESTIONS

Electrochemistry

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Electrochemistry

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Electrochemistry

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Electrochemistry

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Electrochemistry

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Electrochemistry

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Electrochemistry

Consider the following relations for emf of an electrochemical cell : (i) EMF of cell = (Oxidation potential of anode) - (Reduction potential of cathode) (ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode) (iii) EMF of cell = (Reduction potential of anode) + (Reduction potential of cathode) (iv) EMF of cell = (Oxidation potential of anode) - (Oxidation potential of cathode) Which of the above relation(s) is correct?

Electrochemistry

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Electrochemistry

For the cell reaction, $$2{ Ce }^{ 4+ }+Co\rightarrow 2{ Ce }^{ 3+ }+{ Co }^{ 3+ }$$; $${ E }_{ cell }^{ o }$$ cell is $$1.89 V$$. If $${ E }_{ { Co }^{ 3+ }/Co }$$ is $$-0.28 V$$, what is the value of $${ E }_{ { { Ce }^{ 4+ } }/{ { Ce }^{ 3+ } } }^{ o }$$ ?

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