A lead storage cell is discharged which causes the H 2 ​ SO 4 ​ electrolyte from a concentration of 34.6% by mass (density 1.261 g ml −1 at 28 o C) to one of 27% by mass. The original volume of electrolyte is one litre. How many Faraday have left the anode of battery? Note the water is produced by the cell reaction and H 2 ​ SO 4 ​ is used up. Overall reaction is : Pb(s)+PbO 2 ​ +2H 2 ​ SO 4 ​ (l)→2PbSO 4 ​ (s)+2H 2 ​ O

Explanation

To solve the problem of how many Faradays have left the anode of the battery during the discharge of the lead storage cell, we will follow these steps:

### Step 1: Calculate the mass of the solution before discharge
Given that the density of the H₂SO₄ solution is 1.261 g/mL and the volume is 1 liter (1000 mL), we can calculate the mass of the solution.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.261 \, \text{g/mL} \times 1000 \, \text{mL} = 1261 \, \text{g} \]
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