
When a lead storage battery is discharged:
a.) SO$_2$ is evolved
b.) lead sulphate is consumed
c.) lead is formed
d.) sulphuric acid is consumed
Answer
595.5k+ views
Hint: This question is based on the concept of lead-storage cell, or Pb-storage cell in electrochemistry. Write down the reactions for discharge at both the electrodes, i.e. anode, and cathode. Further we can know what happens to storage batteries on discharging.
Complete step-by-step answer:
First, let us write the reaction for the discharge process. It starts acting as a galvanic cell on discharging, and the following mentioned reaction occurs on the negative, and the positive electrode.
The reaction for negative electrode is
Pb(s) + HSO$_4$$^{-}$ + H$_2$O(l) $\rightarrow$ 2e$^{-}$ + PbSO$_4$(s)+H$_3$O$^{+}$(aq)
In this reaction, we can see that the lead reacts with the hydrogen sulphate in the presence of water, and there is formation of lead sulphate with the release of two electrons, and hydronium ions.
The reaction for positive electrode is
PbO$_2$(s) + HSO$_4$$^{-}$ + 3H$_3$O$^{+} $(aq)+ 2e$^{-}$ $\rightarrow$ PbSO$_4$(s)+ 5H$_2$O(l)
Now, we have the second reaction in which the lead oxide reacts with the hydrogen sulphate, and hydronium ion, and there is formation of lead sulphate, and water.
Now, if we analyse the reactions, we can say that the lead sulphate is formed in both the reactions, and there is transfer of two electrons.
Thus, the discharge process going on is due to the conduction of electrons from the negative side to the positive side.
From the reactions, we can also say that the sulphuric acid is lost by the electrolyte, and it becomes water.
Thus, in the end we can conclude that when a lead battery is discharged, sulphuric acid is consumed.
Hence, the correct option is (D).
Additional information: Lead- storage batteries are packed in a thick rubber, or we can say plastic to prevent leakage of the sulphuric acid, as it is corrosive in nature. These batteries should be charged, or changed as the sulphuric acid decomposes, and it requires to be replaced.
Note: There is no need of confusion while writing down the reactions at the negative, and positive electrode. Just remember that the electrons must be transferred, and lead sulphate must be attained as the deposition on the electrodes.
Complete step-by-step answer:
First, let us write the reaction for the discharge process. It starts acting as a galvanic cell on discharging, and the following mentioned reaction occurs on the negative, and the positive electrode.
The reaction for negative electrode is
Pb(s) + HSO$_4$$^{-}$ + H$_2$O(l) $\rightarrow$ 2e$^{-}$ + PbSO$_4$(s)+H$_3$O$^{+}$(aq)
In this reaction, we can see that the lead reacts with the hydrogen sulphate in the presence of water, and there is formation of lead sulphate with the release of two electrons, and hydronium ions.
The reaction for positive electrode is
PbO$_2$(s) + HSO$_4$$^{-}$ + 3H$_3$O$^{+} $(aq)+ 2e$^{-}$ $\rightarrow$ PbSO$_4$(s)+ 5H$_2$O(l)
Now, we have the second reaction in which the lead oxide reacts with the hydrogen sulphate, and hydronium ion, and there is formation of lead sulphate, and water.
Now, if we analyse the reactions, we can say that the lead sulphate is formed in both the reactions, and there is transfer of two electrons.
Thus, the discharge process going on is due to the conduction of electrons from the negative side to the positive side.
From the reactions, we can also say that the sulphuric acid is lost by the electrolyte, and it becomes water.
Thus, in the end we can conclude that when a lead battery is discharged, sulphuric acid is consumed.
Hence, the correct option is (D).
Additional information: Lead- storage batteries are packed in a thick rubber, or we can say plastic to prevent leakage of the sulphuric acid, as it is corrosive in nature. These batteries should be charged, or changed as the sulphuric acid decomposes, and it requires to be replaced.
Note: There is no need of confusion while writing down the reactions at the negative, and positive electrode. Just remember that the electrons must be transferred, and lead sulphate must be attained as the deposition on the electrodes.
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