Question

Rancho being a curious boy and a lover of chemistry wanted to know about the working of this clock battery. He read about it and came to some conclusions. Which of these Rancho’s conclusion are TRUE?

• A. Graphite rod acts as the anode.
• B. The Zinc container as the cathode.
• C. At anode: $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$
• D. At cathode: $Mn{O}_2+N{H}_4^{+}+e^{-}\to MnO\left(OH\right)+N{H}_3$

Answer 1

Answer: (C), (D)

The correct options are
C At anode: $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$
D At cathode: $Mn{O}_2+N{H}_4^{+}+e^{-}\to MnO\left(OH\right)+N{H}_3$

From the last question, we know how it looks like. Now let’s recall about the importance of different things which are present in a Zn – C dry cell.
A dry cell is a compact form of the Leclanche cell. It consists of a cylindrical zinc container which acts as the anode. A graphite rod placed in the centre (but not touching the base) acts as the cathode. The space between the anode and the cathode is so packed that the zinc container is in touch with a metal cap and the cylinder is sealed at the top with pitch. The zinc container is covered with cardboard to protect it from the atmosphere.
The reactions taking place at the electrodes are quite complex. However, they can be represented approximately as follows:
At anode: $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$
At cathode: $Mn{O}_2\left(s\right)+2N{H}_4^{+}\left(aq\right)+2e^{-}\to M{n}_2{O}_3\left(s\right)+2N{H}_3\left(g\right)+H_{2}O$
(i.e., Mn has been reduced from oxidation state +4 to +3)
The $N{H}_3$ formed is not liberated as gas but combines immediately with $Z{n}^{2+}$ions and $C{l}^{-}$ ions to form complex salt $\left[Zn\right(N{H}_3{)}_4{]}^{2+}C{l}_2^{2-}$ (tetraammine zinc (II) chloride). These cells have voltage in the range 1.25 V to 1.50 V. However, they do not have a long life because the acidic $N{H}_{4}Cl$ corrodes the zinc container even when the cell is not in use.