A galvanic (voltaic) cell contains a copper cathode immense in a copper (II) chl

Question

A galvanic (voltaic) cell contains a copper cathode immense in a copper (II) chloride solution and a nickel anode immersed in a nickel (II) chloride solution. The two solutions are connected with a salt bridge. Write the balanced equation for the galvanic cell. Phases are optional. A current of 2.43 A is observed flowing through the cell for a period of 1.79 hours. Calculate the amount of charge flowing through the circuit during this time. Calculate the number of moles of electrons flowing through the circuit during this time Calculate the mass change in grams of the copper and nickel electrodes

Explanation / Answer

Cu/CuCl2(aq.) cathode

Ni/NiCl2(aq.) anode

(i)

Balanced redox reaction is ,

Cu2+ (aq.) + Ni (s) ----------> Ni2+ (aq.) + Cu (s)

(ii)

Amount of charge = current in amp * time in seonds

Q = 2.43 * (1.79*60*60)

Q = 15659 amp.s

Q = 15659 C

(iii)

1 Faraday of charge (96485 C) = 1 mol of electrons

then, 15659 C of charge = 1 * 15659 / 96485 = 0.162 mol of electrons.

(iv) Cu2+ (aq.) + 2 e ----------> Cu (s)

From this equation, 2 Faradays's i.e 2 * 96485 C of charge can deposit 1 mol i.e. 63.5 g. of Cu

Then, 15659 C of charge can deposit 15659 * 63.5 / ( 2 * 96485 ) = 5.15 g. of Cu

So, cathode electrons gains the mass of 5.15 g.

Similarly, Ni (s) ---------> Ni2+ (aq.) + 2 e

SO, to produce 2 Faraday's i.e 2 * 96485 C of charge 1 mol i.e 58.69 g. of Ni is to be consumed

then to produce 15659 C of charge 15659 * 58.69 / (2*96485) = 4.76 g. of Ni is to be consumed.

So, anode electrode loses a mass of 4.76 g.

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