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NCERT Exemplar · Q40

Q.Two Daniell-type cells, Cell 'A' and Cell 'B', are connected to each other in opposition (their EMFs oppose). In each cell a zinc electrode dips in a Zn2+(aq) solution and a copper electrode dips in a Cu2+(aq) solution, the two half-cells being joined by a salt bridge. When two such cells are coupled in opposition, the cell of higher EMF discharges and drives the cell of lower EMF in reverse, so the lower-EMF cell is forced to behave as an electrolytic cell. Answer the following.

(i) Cell 'A' has ECell=2 VE_{Cell} = 2\ \text{V} and Cell 'B' has ECell=1.1 VE_{Cell} = 1.1\ \text{V}. Which of the two cells, 'A' or 'B', will act as an electrolytic cell, and which electrode reactions will occur in that cell?
(ii) If Cell 'A' has ECell=0.5 VE_{Cell} = 0.5\ \text{V} and Cell 'B' has ECell=1.1 VE_{Cell} = 1.1\ \text{V}, what will be the reactions at the anode and the cathode of the cell that is forced to act as the electrolytic cell?
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When two cells are joined in opposition, the one with the larger EMF acts as a galvanic cell (source) and forces the smaller-EMF cell to run backwards, i.e. as an electrolytic cell. Running a Daniell cell backwards reverses its normal reactions: the zinc electrode becomes the cathode (Zn2+ deposits as Zn) and the copper electrode becomes the anode (Cu dissolves as Cu2+).

Concept

A Daniell cell spontaneously (as a galvanic cell) undergoes:

  • Anode (Zn): Zn → Zn2+ + 2e-
  • Cathode (Cu): Cu2+ + 2e- → Cu

If an external source of higher EMF pushes current through it the opposite way, the cell is electrolysed and every electrode reaction reverses:

  • Zn electrode becomes the cathode: Zn2+ + 2e- → Zn
  • Cu electrode becomes the anode: Cu → Cu2+ + 2e-

(i) EA=2 VE_A = 2\ \text{V}, EB=1.1 VE_B = 1.1\ \text{V}

  1. Compare the EMFs: EA(2 V)>EB(1.1 V)E_A (2\ \text{V}) > E_B (1.1\ \text{V}).
  2. The stronger cell, A, works as a galvanic cell and supplies current; the weaker cell, B, is driven in reverse and therefore behaves as the electrolytic cell.
  3. Reactions in Cell B (reversed Daniell):
    • Cathode (zinc electrode): Zn2+ + 2e- → Zn (reduction)
    • Anode (copper electrode): Cu → Cu2+ + 2e- (oxidation)

(ii) EA=0.5 VE_A = 0.5\ \text{V}, EB=1.1 VE_B = 1.1\ \text{V} …

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