A rod of metal A is dipped in a 1 M aqueous solution of B2+ ions at 25 °C. The standard electrode potentials are E°(A2+/A) = −0.76 V and E°(B2+/B) = +0.34 V. What happens?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B will gradually deposit on A
The lower the standard electrode potential, the stronger the metal is as a reducing agent. A metal with a lower E° reduces the ions of a metal with a higher E°.
Here E°(A2+/A) = −0.76 V is lower than E°(B2+/B) = +0.34 V, so metal A gives electrons to the B2+ ions:
\( \displaystyle \ce{A(s) + B^2+(aq) -> A^2+(aq) + B(s)} \)
\( \displaystyle E^{\circ}=E^{\circ}_{\ce{B^2+/B}}-E^{\circ}_{\ce{A^2+/A}}=0.34-(-0.76)=1.10\,\mathrm{V} \)
E° is positive, so the reaction takes place on its own. A dissolves as A2+, and the B2+ ions are reduced to metal B, which deposits on the rod of A. These are the values of zinc and copper, so this is the reaction of the Daniell cell.
Why the other options are incorrect
Option (1) A will gradually deposit on the rod — Incorrect. A is oxidised and goes into solution as A2+; the rod loses metal A, it does not gain it.
Option (3) No reaction will occur — Incorrect. No reaction would mean A has the higher E°. Here E° for the reaction is +1.10 V, so it is spontaneous.
Option (4) Water will be decomposed into H2 and O2 — Incorrect. Splitting water into H2 and O2 is a non-spontaneous change that needs electrical energy (electrolysis); a metal rod dipped in a salt solution does not do it.
NCERT (2026-27 edition):
2.2.1 Measurement of Electrode Potential – “… The overall reaction of the cell is the sum of above two reactions and we obtain the equation:”
\( \displaystyle \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \)
\( \displaystyle \text{emf of the cell}=E^{\circ}_{\text{cell}}=E^{\circ}_{R}-E^{\circ}_{L}=0.34\,\mathrm{V}-(-0.76)\,\mathrm{V}=1.10\,\mathrm{V} \)
“… It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction.”
Table 2.1: Standard Electrode Potentials at 298 K
