A standard galvanic cell is constructed with Cd 2+ | Cd and Al 3+ | Al half cell

Question

A standard galvanic cell is constructed with Cd2+|Cd and Al3+|Al half cell compartments connected by a salt bridge. Which of the following statements are correct?

Hint: Refer to a table of standard reduction potentials.

(Choose all that apply.)

---In the external circuit, electrons flow from the Al3+|Al compartment to the Cd2+|Cd compartment.

---The anode compartment is the Al3+|Al compartment.

---The cathode compartment is the Al3+|Al compartment.

---Cd is oxidized at the anode.

---As the cell runs, anions will migrate from the Al3+|Al compartment to the Cd2+|Cd compartment.

A standard galvanic cell is constructed with Cr3+ | Cr2+ and Fe2+ | Fe half cell compartments connected by a salt bridge. Which of the following statements are correct?

Hint: Refer to a table of standard reduction potentials.

(Choose all that apply.)

---Fe is oxidized at the anode.

---The anode compartment is the Fe2+|Fe compartment.

---Fe2+ is reduced at the cathode.

---In the external circuit, electrons flow from the Fe2+|Fe compartment to the Cr3+|Cr2+ compartment.

---As the cell runs, anions will migrate from the Fe2+|Fe compartment to the Cr3+|Cr2+ compartment.

Explanation / Answer

Solution:- (1) The reduction equations for the two given electrodes are as given below...

Cd2+(aq) + 2e- -------> Cd(s) E0 = -0.40 V

Al3+(aq) + 3e- --------> Al(s) E0 = -1.66 V

Standard reduction potential for Aluminum electrode is more negative it means the oxidation of Aluminum and reduction of cadmium would take place. The over all cell reaction is...

2Al(s) + 3Cd2+(aq) -------> 2Al3+(aq) + 3Cd(s)

Oxidation takes place at anode and reduction at cathode. So, the anode is Al3+/Al and cathode is Cd2+/Cd.

In external circuit the electrons flow from anode to cathode.

At anode, Al dissolves into the solutions as Al3+, so to maintain the neutrality, negative ions from salt bridge moves towards anode that is Al3+/Al.

So, the correct statements are------

(i) In the external circuit, electrons flow from the Al3+|Al compartment to the Cd2+|Cd compartment.

(ii) The anode compartment is the Al3+|Al compartment.

(2) The two reducttions half equatons are...

Cr3+(aq) + 1e- ----> Cr2+(aq) E0 = -0.407 V

Fe2+(aq) + 2e- ------> Fe(s) E0 = -0.410 V

Stanrad reduction potential for Fe half cell is slighly greater than Cr half cell. So, oxidation of Fe and reduction of Cr would be taking place. Anode is of Iron and cathode is of chromium. Electrons in external circuit flow from anode to cathode. So, the correct statements are...

(i)  Fe is oxidized at the anode.

(ii) The anode compartment is the Fe2+|Fe compartment.

(iii) In the external circuit, electrons flow from the Fe2+|Fe compartment to the Cr3+|Cr2+ compartment.

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