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I made a dry cell battery from zinc (Zn), carbon (C) and ammonium chloride (NH4Cl) which produced a current of 0.4 amps and voltage of 1.1V. However when I could not find the redox reaction that was occurring.

What is the chemical reaction happening in the cell?

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    $\begingroup$ If you put litmus paper near the carbon electrode, when your voltaic cell is in operation, does the litmus paper turn blue? In other words, is the electrolyte basic near the carbon cathode? $\endgroup$ – Ed V Mar 17 '20 at 2:43
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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 : $$\ce{Zn + 4 NH_4^+ -> [Zn(NH_3)_4]^{2+} + 4 H+ + 2 e-}\label{rxn:zn4}\tag{1}$$ 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 $\ce{E_2° = - 1.03 V}$. $$\ce{Zn + 4 NH3 -> [Zn(NH_3)_4]^{2+} + 2 e-}\label{rxn:zn}\tag{2}$$ 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 : $$\ce{2 H2O + 2 e- -> 2 OH- + H2}\label{rxn:z}\tag{4}$$ 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 : $$\ce{2 H+ + 2 e- -> H2}\label{rxn:znn}\tag{5}$$ 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 "Chemistry Data Book" from J. G. Stark and H. G. Wallace, John Murray, London, $1994$, Table $43$, p. $70$.

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Per a source, the chemical reactions in a zinc–carbon battery can be detailed as follows, to quote:

In a zinc–carbon dry cell, the outer zinc container is the negatively charged terminal. The zinc is oxidised by the charge carrier, chloride (Cl−) via the following half reactions:

Anode (oxidation reaction, marked −)

$\ce{Zn + 2 Cl− → ZnCl2 + 2 e−}$

Cathode (reduction reaction, marked +)

$\ce{2 MnO2 + 2 NH4Cl + H2O + 2 e- → Mn2O3 + 2 NH4OH + 2 Cl−}$

Other side reactions are possible, but the overall reaction in a zinc–carbon cell can be represented as

$\ce{Zn + 2 MnO2 + 2 NH4Cl + H2O → ZnCl2 + Mn2O3 + 2 NH4OH}$

The battery cell is claimed to have an electromotive force (e.m.f.) of about 1.5 V. Note, personally above, I would write NH4OH as NH3(aq).

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    $\begingroup$ The cell that was made doesn't have MnO2 in it. $\endgroup$ – AEON Mar 16 '20 at 9:20
  • $\begingroup$ Anyway, $\ce{ NH_4OH}$ does not exist. And I would write the anodic reaction simply as $\ce{Zn -> Zn^{2+} + 2 e^-}$. Further more I would not introduce the molecule $\ce{NH_4Cl}$, but $\ce{NH_4^+}$ ion. $\endgroup$ – Maurice Apr 15 '20 at 10:42

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