Question

Question: A wire is fastened across the terminals of the Leclanche cell in the figure. Indicate the direction of election flow in the wire.

Short answer

Answer:

From the figure of the Lechlanche cell, the zinc cell serves as the anode whereas the axial graphite rod serves as the cathode. For proper ion movement, we are provided with Ammonium chloride and zinc chloride inside the cell. The direction of flow of current is shown in the image below.

Step-by-step solution

Step-1: The reactions involved at cathode and anode.

The reactions involved at cathode and anode.

Reaction taking place at cathode:

$$\begin{gathered} 2N{H_4}^{+}+2e^{-}\to H_2\left(g\right)+2N{H}_3\left(aq\right) \\ {H}_2\left(g\right)+2Mn{O}_2\left(s\right)\to M{n}_2{O}_3\left(s\right)+H_{2}O\left(l\right) \end{gathered}$$

Reaction taking place at anode:

$Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$

Step-2: Deduce the overall cell reaction.

The overall cell reaction after joining the reactions at both anode and cathode comes out to be:

$$\begin{gathered} 2N{H_4}^{+}+2e^{-}\to H_2\left(g\right)+2N{H}_3\left(aq\right) \\ {H}_2\left(g\right)+2Mn{O}_2\left(s\right)\to M{n}_2{O}_3\left(s\right)+H_{2}O\left(l\right) \\ Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-} \\ {\text{Zn(s) + 2MnO}}_{\text{2}}\text{(s) + 2N}{{\text{H}}_{\text{4}}}^{\text{+}}\text{(aq)}\to {\text{Zn}}^{\text{2 +}}{\text{(aq) + Mn}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 2NH}}_{\text{3}}{\text{(aq) +H}}_{\text{2}}\text{O(l)} \end{gathered}$$

Step-3: To determine the electron flow.

The electron gets produced during the oxidation at the anode.

From the overall reaction, the zinc gets oxidised and thus the electron flows from zinc shell to the graphite cell.

Step-4: To show electron movement.

The figure above shows the direction of electron flow.

Hence, the electron produced due to oxidation in the anode flows from zinc shell to graphite shell.