Reaction in Zinc-Carbon-Ammonium Chloride battery
Surprisingly this battery uses Zn without any oxidant like $\ce{MnO_2}$ and it  produces a voltage as high as $1.1$ V. In order to establish its characteristics, the two compartments should be studied independently.
First, the anode, with zinc. $\ce{Zn}$ should not produce simple $\ce{Zn^{2+}}$ ions, as there are ammonia and ammonium chloride in the solution. So the anode reaction is probably :  $$\ce{Zn  + 4 NH_4^+ -&gt;  [Zn(NH_3)_4]^{2+}  +  4 H+  +  2 e-}\label{rxn:zn4}\tag{1}$$ The reduction potential for this anodic equation is not given in the tables. But it may be obtained from the next equation, whose standard reduction potential is known to be $\ce{E_2° = - 1.03 V}$.  $$\ce{Zn  + 4 NH3 -&gt;  [Zn(NH_3)_4]^{2+}  +  2 e-}\label{rxn:zn}\tag{2}$$  In a $1$ M solution of $\ce{NH4Cl}$, $\ce{[NH_3]}$ = $2.4 ·10^{-5}$ M. So the corresponding reduction potential of ($2$) is : $$\ce{E_2 = - 1.03 \mathrm{V} + 0.0296\mathrm{V}· log (2.4 ·10^{-5})^4} = - 1.03~ \mathrm{V} - 0.54~ \mathrm{V}  = -1.57 ~\mathrm{V}\label{rxn:nh}\tag{3}$$
Now the cathodic reaction should be studied. Suppose the cathode reaction is  : $$\ce{2 H2O  + 2 e- -&gt; 2 OH- + H2}\label{rxn:z}\tag{4}$$ with a redox potential equal to  $- 0.83$ V. Couping these two expressions ($2$) and ($3$) are never sufficient to get $-1.1$ V as the overall voltage. So, as the ammonium chloride solution is acidic, the cathode reaction must rather be : $$\ce{2 H+ + 2 e- -&gt;  H2}\label{rxn:znn}\tag{5}$$ In a $1$ M  $\ce{NH4Cl}$ solution, the pH is $4.62$, so the corresponding potential of the cathode is E$\ce{_5}$ = - $0.059$ pH = - $0.059·4.62$ V = - $0.27$ V.
To summarize, the anode reaction is ($1$) or ($2$), at will, and the cathode rection is ($5$). As a consequence, the theoretical voltage of the cell is : $\ce{E_5}$ - $\ce{E_2}$ = - $0.27$ V + $1.57$ V = $1.3$ V, if the concentration of ammonium chloride is $1$ M.
Ref.: Standard electrode potentials have been taken from &quot;Chemistry Data Book&quot; from J. G. Stark and H. G. Wallace, John Murray, London, $1994$, Table $43$, p. $70$.