Question

Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	Electrochemical equivalent	(i)	Potential difference x Quantity of charge
(B)	Faraday	(ii)	Mass of substance deposited by one coulomb of charge
(C)	Ampere	(iii)	Charge carried by one mole of electrons
(D)	Electrical energy	(iv)	One coulomb of electric charge passed through one second

• A. (A)$\to$(ii), (B)$\to$ (i), (C)$\to$ (iii), (D)$\to$ (iv)
• B. (A)$\to$ (ii), (B)$\to$ (iii), (C)$\to$ (iv), (D)$\to$ (i)
• C. (A)$\to$ (iii), (B)$\to$ (iv), (C)$\to$ (i), (D)$\to$ (ii)
• D. (A)$\to$ (iv), (B)$\to$ (i), (C)$\to$ (ii), (D)$\to$ (iii)

Answer 1

The correct option is B (A)$\to$ (ii), (B)$\to$ (iii), (C)$\to$ (iv), (D)$\to$ (i)

• Electrochemical equivalent: The weight of a substance deposited or evolved during electrolysis by the passage of a specified quantity of electricity and usually expressed in grams per coulomb
  
• Faraday: Ameasureoftheelectricchargecarriedbyonemoleofelectrons,usedin electrolysisasthe quantity ofcharge requiredtodepositorliberateonegramequivalentweightofa substance.
• Ampere: By definition, if 1 Coulomb of Charge flows through a Conductor in 1 second then the magnitude of the current is said to be 1 A.
• Electrical Energy: Electrical energy is the work done by electric charge. If current $i$ ampere flows through a conductor or through any other conductive element of potential difference V volts across it, for time $t$ second. We know $Q=I\times t$ and hence electrical energy becomes, Potential difference$\times$quantity of charge