Question

Question has to do with electrochemistry. Very confused on this concept and would greatly appreciate if you could explain your answers!

3. Consider the cell at 298 K Zn (s) | Zn2 (1.00 M) || Cu2 (1.00 M) | Cu (s) for which E following changes are made? 1.10 V. Does the cell voltage increase, decrease, or remain the same when each of the (a) The cell runs for 5 minutes. (b) Excess 1.0 M ammonia is added is added to the cathode compartment (c) H2S gas is bubbled into the anode compartment

Answer 1

$Zn(s) +Cu^{2+} (aq )\to Cu(s)+Zn^{2+} (aq)$

$E_{cell}=E_{cell}^0-\frac{0.0592}{n}log_{10}Q$

$E_{cell}=1.10 \ V-\frac{0.0592}{2}log_{10}\frac{[Zn^{2+}]}{[Cu^{2+}]}$

(a) The cell runs for minutes. Some Cu(II) ions are reduced to Cu metal. Due to this, Cu(II) ion concentration decreases. Some Zn solid is converted to Zn(II) ions. Hence, concentration of Zn(II) ions increases. Hence, $Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$ increases and $log_{10}Q$ increases. Hence, cell voltage decreases.

(b) Excess 1.0 M ammonia is added to cathode compartment.

Cathode is $Cu|Cu^{2+}$

Ammonia reacts with Cu(II) ions to form complex ion.

$Cu^{2+}(aq) + 4 NH_3(aq) \to [Cu(NH_3)_4]^{2+} (aq)$

In the above reaction, Cu(II) ions are consumed. Due to this concentration of free Cu(II) ions decreases.

Hence, $Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$ increases and $log_{10}Q$ increases. Hence, cell voltage decreases.

(c)

H2S gas is passed through anode solution.

Anode is $Zn|Zn^{2+}$

Zn(II) ions react with H2S to form zinc sulphide precipitate.

$Zn^{2+}+H_2S \to 2H^++ZnS \downarrow$

In the above reaction, Zn(II) ions are consumed. Due to this concentration of free Zn(II) ions decreases.

Hence, $Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$ decreases and $log_{10}Q$ decreases. Hence, cell voltage increases.