Question

A lead storage cell is discharged which causes the ${H_2}S{O_4}$ electrolyte from a concentration of 34.6% by mass (density = 1.261 $g.m{l^{ - 1}}$ at 28$^ \circ C$) to one of 27% by mass. The original volume of electrolyte is one litre. How many Faraday have left the anode of the battery? Note the water is produced by the cell reaction and ${H_2}S{O_4}$ is used up. Overall reaction is:
                       $Pb(s) + Pb{O_2} + 2{H_2}S{O_4}(l) \to 2PbS{O_4}(s) + 2{H_2}O$
(A) 2.5 F
(B) 0.22 F
(C) 1.1 F
(D) 0.97 F

Answer 1

Hint: First find out the amount of solution, ${H_2}S{O_4}$ and ${H_2}O$ before the discharge. Consider the amount of ${H_2}O$ produced by x g which is equal to ${H_2}S{O_4}$ consumed. Then find out the amount of ${H_2}S{O_4}$left after discharge. Final conc. of ${H_2}S{O_4}$ is 27%, so use the mass % formula to find out the value of x and use it to calculate the moles of ${H_2}O$ produced. Total charge produced is equal to nF.

Complete step by step answer:
-First of all we will find out the weight of electrolyte and ${H_2}S{O_4}$ (34.6% by mass) in it present before discharge at the anode. The initial volume of the electrolyte is 1L (1000ml) and its density is = 1.261 $g.m{l^{ - 1}}$
So, weight of the solution before discharge = V × d
                                                                              = 1000 × 1.261
                                                                             = 1261g

Since ${H_2}S{O_4}$ concentration in it is 34.6% by mass, its weight will be:
                                    = $\dfrac{{34.6}}{{100}} \times 1261$
                                    = 436.306g
So, the amount of water would be = total weight of solution – weight of ${H_2}S{O_4}$
                                                              = 1261 – 436.306
                                                              = 824.694g
-After discharge occurs at the anode more ${H_2}O$ is produced and concentration of ${H_2}S{O_4}$reduces to 27% by mass. Let the amount of ${H_2}O$ produced as a result of the net reaction be x g.
                      $Pb(s) + Pb{O_2} + 2{H_2}S{O_4}(l) \to 2PbS{O_4}(s) + 2{H_2}O$

So, the moles of ${H_2}O$ produced will be = $\dfrac{x}{{18}}$ mol
Also from the net reaction we can see that: moles of ${H_2}O$ produced is equal to moles of ${H_2}S{O_4}$ consumed = $\dfrac{x}{{18}}$ mol
So, the weight of ${H_2}S{O_4}$ consumed = moles × Mol. Wt.
                                                                        = $\dfrac{x}{{18}} \times 98$ g
After consumption, the amount of ${H_2}S{O_4}$ left would be:
                                                                        = $436.306 - \dfrac{{98x}}{{18}}$ g

-Now we will find out the total amount of electrolyte solution after the discharge occurs:
 Final wt. of electrolyte = (Initial wt.) – (wt. of ${H_2}S{O_4}$ consumed) + (wt. of ${H_2}O$ produced)
                                         = $1261 - \dfrac{{98x}}{{18}} + x$

-Now the question says that the final concentration by mass of ${H_2}S{O_4}$ is 27%. We also know that we calculate mass by percentage using formula:
                     $\% mass = \dfrac{{wt.{H_2}S{O_4}left}}{{wt.solution}} \times 100$

We will now use the above formula to find out the value of variable ‘x’ which was the amount of ${H_2}O$ produced.

  $27 = \left[ {\dfrac{{436.306 - \dfrac{{98x}}{{18}}}}{{1261 - \dfrac{{98x}}{{18}} + x}}} \right] \times 100$
                                         0.27 = $\dfrac{{7853.508 - 98x}}{{22698 - 80x}}$
                0.27 (22698 – 80x) = 7853.508 – 98x
                     6128.46 – 21.6x = 7853.508 – 98x
                              98x – 21.6x = 7853.508 – 6128.46
                                         76.4x = 1725.048
                                                 x = 22.579 g

So, the amount of ${H_2}O$ produced is 22.579 g.
Moles of ${H_2}O$ produced = $\dfrac{{22.579}}{{18}}$ = 1.254 mol

-We will now calculate the total charge produced at the anode.
The formula for calculating total charge = nF ; and the value of n= 1.254 moles
                            Q = n F
                           = 1.254 F

  So, we can now tell that the amount of Faraday that left the anode is 1.254 F.
The correct option would be: (C) 1.1 F
 So, the correct answer is “Option C”.

Note: To calculate the charge we use moles of ${H_2}O$ produced and not ${H_2}S{O_4}$ consumed, although weight of both is same (because we need to find faradays that left the anode).