Write a note on lead storage cell (battery).
(N/A) The most important secondary cell is the lead storage battery. It can be recharged after its discharge.
Construction:
Anode: It consists of a grid of lead packed with spongy lead $(Pb)$.
Cathode: $A$ grid of lead packed with lead dioxide $(PbO_2)$.
Electrolyte: $A$ $38\%$ solution of sulphuric acid $(H_2SO_4)$ is used as an electrolyte.
(Density: initial $1.25$ to $1.30 \ g \ mL^{-1}$)
Discharging reaction of the cell:
When the cell is in working condition,the following reactions are carried out:
Anode: Oxidation of $Pb$ into $PbSO_4$.
$(i) \ Pb_{(s)} + SO_{4(aq)}^{2-} \rightarrow PbSO_{4(s)} + 2e^{-}$
Cathode: Reduction of $PbO_2$ into $PbSO_4$.
$(ii) \ PbO_{2(s)} + SO_{4(aq)}^{2-} + 4H^+_{(aq)} + 2e^- \rightarrow PbSO_{4(s)} + 2H_2O_{(l)}$
Overall reaction: Adding $(i)$ and $(ii)$ gives the overall reaction of the lead storage cell:
$(iii) \ Pb_{(s)} + PbO_{2(s)} + 2H_2SO_{4(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)}$
Construction:
Anode: It consists of a grid of lead packed with spongy lead $(Pb)$.
Cathode: $A$ grid of lead packed with lead dioxide $(PbO_2)$.
Electrolyte: $A$ $38\%$ solution of sulphuric acid $(H_2SO_4)$ is used as an electrolyte.
(Density: initial $1.25$ to $1.30 \ g \ mL^{-1}$)
Discharging reaction of the cell:
When the cell is in working condition,the following reactions are carried out:
Anode: Oxidation of $Pb$ into $PbSO_4$.
$(i) \ Pb_{(s)} + SO_{4(aq)}^{2-} \rightarrow PbSO_{4(s)} + 2e^{-}$
Cathode: Reduction of $PbO_2$ into $PbSO_4$.
$(ii) \ PbO_{2(s)} + SO_{4(aq)}^{2-} + 4H^+_{(aq)} + 2e^- \rightarrow PbSO_{4(s)} + 2H_2O_{(l)}$
Overall reaction: Adding $(i)$ and $(ii)$ gives the overall reaction of the lead storage cell:
$(iii) \ Pb_{(s)} + PbO_{2(s)} + 2H_2SO_{4(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)}$
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View SolutionThe standard reduction potential data at $25^{\circ} C$ is given below.
$E^{\circ}(Fe^{3+}, Fe^{2+}) = +0.77 \ V$
$E^{\circ}(Fe^{2+}, Fe) = -0.44 \ V$
$E^{\circ}(Cu^{2+}, Cu) = +0.34 \ V$
$E^{\circ}(Cu^{+}, Cu) = +0.52 \ V$
$E^{\circ}(O_{2(g)} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O) = +1.23 \ V$
$E^{\circ}(O_{2(g)} + 2H_{2}O + 4e^{-} \rightarrow 4OH^{-}) = +0.40 \ V$
$E^{\circ}(Cr^{3+}, Cr) = -0.74 \ V$
$E^{\circ}(Cr^{2+}, Cr) = -0.91 \ V$
Match $E^{\circ}$ of the redox pair in List-$I$ with the values given in List-$II$ and select the correct answer using the code given below the lists:
List-$I$ List-$II$ $P. \ E^{\circ}(Fe^{3+}, Fe)$ $1. \ -0.18 \ V$ $Q. \ E^{\circ}(4H_{2}O \rightleftharpoons 4H^{+} + 4OH^{-})$ $2. \ -0.4 \ V$ $R. \ E^{\circ}(Cu^{2+} + Cu \rightarrow 2Cu^{+})$ $3. \ -0.04 \ V$ $S. \ E^{\circ}(Cr^{3+}, Cr^{2+})$ $4. \ -0.83 \ V$
Codes: $P \quad Q \quad R \quad S$
$E^{\circ}(Fe^{3+}, Fe^{2+}) = +0.77 \ V$
$E^{\circ}(Fe^{2+}, Fe) = -0.44 \ V$
$E^{\circ}(Cu^{2+}, Cu) = +0.34 \ V$
$E^{\circ}(Cu^{+}, Cu) = +0.52 \ V$
$E^{\circ}(O_{2(g)} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O) = +1.23 \ V$
$E^{\circ}(O_{2(g)} + 2H_{2}O + 4e^{-} \rightarrow 4OH^{-}) = +0.40 \ V$
$E^{\circ}(Cr^{3+}, Cr) = -0.74 \ V$
$E^{\circ}(Cr^{2+}, Cr) = -0.91 \ V$
Match $E^{\circ}$ of the redox pair in List-$I$ with the values given in List-$II$ and select the correct answer using the code given below the lists:
| List-$I$ | List-$II$ |
| $P. \ E^{\circ}(Fe^{3+}, Fe)$ | $1. \ -0.18 \ V$ |
| $Q. \ E^{\circ}(4H_{2}O \rightleftharpoons 4H^{+} + 4OH^{-})$ | $2. \ -0.4 \ V$ |
| $R. \ E^{\circ}(Cu^{2+} + Cu \rightarrow 2Cu^{+})$ | $3. \ -0.04 \ V$ |
| $S. \ E^{\circ}(Cr^{3+}, Cr^{2+})$ | $4. \ -0.83 \ V$ |
Codes: $P \quad Q \quad R \quad S$
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