A cell is drawn as two beakers: on the left a zinc rod dips into 1 M ZnSO4 solution, on the right a copper rod dips into 1 M CuSO4 solution, a salt bridge joins the two solutions and a wire through a voltmeter joins the rods. E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V at 298 K. Which statement about this cell is correct?
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This is the Daniell cell. Zinc, with the lower reduction potential, is oxidised at the anode and its electrons travel through the wire to the copper cathode, where Cu²⁺ is reduced. E°cell = 0.34 − (−0.76) = 1.10 V.
- A.
Zinc has the lower reduction potential, so it is the anode; 0.42 V wrongly adds the two potentials.
- B.
emf = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V, not the sum −0.76 + 0.34.
- D.
Zinc is oxidised at the anode and releases electrons, so they flow from zinc to copper.
NCERT: Class 12 Chemistry, Chapter 2