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NCERT Exemplar · Q41

Q.A galvanic cell is set up as follows: a zinc plate dips in a Zn2+(aq) solution in beaker I, and a silver plate dips in an Ag+(aq) solution in beaker II. The two beakers are joined by a salt bridge, and the zinc and silver plates are connected through an external wire (carrying a switch/meter). Given the standard reduction potentials EZn2+∣Zn∘=−0.76 VE^{\circ}_{Zn^{2+}\mid Zn} = -0.76\ \text{V} and EAg+∣Ag∘=+0.80 VE^{\circ}_{Ag^{+}\mid Ag} = +0.80\ \text{V}, answer the following.

(i) State the direction of electron flow in the external wire.
(ii) Is the silver plate the anode or the cathode?
(iii) What will happen if the salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will the concentrations of Zn2+ ions and Ag+ ions be affected while the cell functions?
(vi) How will the concentrations of Zn2+ ions and Ag+ ions be affected after the cell becomes 'dead'?
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This is a Zn-Ag galvanic cell. Because EAg+∣Ag∘(+0.80 V)>EZn2+∣Zn∘(−0.76 V)E^{\circ}_{Ag^{+}\mid Ag}(+0.80\ \text{V}) > E^{\circ}_{Zn^{2+}\mid Zn}(-0.76\ \text{V}), zinc is oxidised at the anode and silver ions are reduced at the cathode. The overall reaction is Zn + 2Ag+ → Zn2+ + 2Ag. Electrons travel through the wire from zinc to silver; [Zn2+] grows and [Ag+] shrinks until the cell reaches equilibrium and goes dead.

Concept and electrode assignment

The electrode with the lower (more negative) reduction potential is oxidised (anode); the one with the higher (more positive) reduction potential is reduced (cathode).

  • Anode (zinc): Zn → Zn2+ + 2e-
  • Cathode (silver): Ag+ + e- → Ag (×2 to balance electrons)
  • Overall: Zn + 2Ag+ → Zn2+ + 2Ag, with Ecell∘=0.80−(−0.76)=1.56 VE^{\circ}_{cell} = 0.80 - (-0.76) = 1.56\ \text{V}.

(i) Direction of electron flow

Electrons are released at the zinc (anode) and consumed at the silver (cathode). In the external wire they flow from the zinc plate to the silver plate (conventional current flows the opposite way, silver → zinc).

(ii) Silver plate — anode or cathode?

Silver, having the higher reduction potential, is where reduction (Ag+ + e- → Ag) occurs, so the silver plate is the cathode (the positive electrode).

(iii) If the salt bridge is removed

The salt bridge completes the circuit and keeps each half-cell electrically neutral. If it is removed, the internal ionic path is broken: charge builds up (the anode compartment becomes positively charged as Zn2+ forms, the cathode compartment negatively charged as Ag+ is removed). This opposing charge stops the electrode reactions, so the cell stops functioning and no current flows.

(iv) When does the cell stop functioning? …

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