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emf of a chemical cell depends on...

emf of a chemical cell depends on

A

Distance between electrodes

B

Area of electrods

C

Nature of electrolyte

D

Chemical potential of electrons

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Consider the following relations for emf of a 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 realtions are correct?

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AAKASH INSTITUTE ENGLISH-CURRENT ELECTRICITY-ASSIGNMENT(SECTION-A(OBJECTIVE TYPE QUESTIONS))
  1. Effective resistance across AB in the network shown in

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  2. A uniform wire os resistance 12 Omega is bent to form a circle. Effec...

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  3. A wire of resistance R is cut into ‘ n ’ equal parts. These parts are ...

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  4. Potential difference across AB in the circuit as shown in figure.

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  5. Potential difference across AB in the network shown is

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  6. emf of a chemical cell depends on

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  7. When current supplied by a cell to a circuit is 0.3 A, its terminal po...

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  8. Potential difference across AB in the networ shown is

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  9. Find the current through the 10(Omega)resistor shown in figure.

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  10. Three identical cells are connected in parallel across AB. Net acr...

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  11. Five cells each of emf E and internal resistance r are connecte in se...

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  12. Potentail difference V(A) - V(B) in the network shown in

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  13. Potential difference across AB , i.e., V(A)-V(B) is

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  14. Potential difference V(B)-V(A) in the network shown is

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  15. Current l in the network shown in figure is

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  16. Value of the resistance R in the figure is

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  17. Current through resistor R3 as shown in figure is . Given that R1=10 ...

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  18. Resistance across AB as shown in figure is

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  19. Determination of unknonw resistance in metre bridge is most accurate ,...

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  20. Two cells of emf E(1) and E(2) ( E(1) gt E(2)) are connected in series...

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