Question

A steady current of 100 A is passed through a cell containing AgNO3 solution connected in series with a cell containing dilute H2SO4. This current results in deposition of 1.08 g at cathode.(At mass of Ag is equal to 108) 1. for how long the current was passed? 2. calculate the volume of O2 gas evolved at STP in the cell with dilute H2SO4.​

Answer 1

Answer:

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Answer 2

1. The current was passed for 9.65 seconds.

2. The volume of O₂ gas evolved at STP is 56 ml or 0.056 l.

Explanation:

It is given that:

• Quantity of current passed through the cell (I) = 100 A

• Amount of deposition at the cathode (W) = 1.08 g

• The Atomic mass of Ag = 108 g/mol

Since the standard reduction potential of Ag is greater than the standard reduction potential of H, the following reactions take place-

• At anode,

$H\rightarrow H^{+} + e$

• At cathode,

$Ag^{+} + e \rightarrow Ag(s)$

• We know that $W = \frac{Eq.wt}{96500} \times Q$

Rearranging the above equation,

$Q = \frac{W\times96500}{Eq.wt}$

In this problem Equivalent weight of Ag = Atomic mass of Ag, therefore,

$Q = \frac{1.08 \times 96500}{108}$

$Q = 965 C$

• To deposit 1.08 g of Ag at the cathode, the amount of charge required (Q) is 965 C

1. By definition of current, we get the formula,

$I=\frac{Q}{t}$

or

$t = \frac{Q}{I}$

$t = \frac{965}{100} = 9.65 s$

• The current was passed for 9.65 seconds to deposit 1.08 g of Ag at the cathode.

2. To find the volume of O₂ evolved when 965 C of charge is passed through the cell:

$2H_{2}O(l) \rightarrow 4H^{+}(aq)+O_{2}(g)+4e$

• No. of moles of O₂ formed when 965 C of charge is passed through the cell (n) is:

$n=\frac{W}{At.wt} =\frac{1}{4\times 96500} \times 965$

$n=0.0025 moles$

• 1 mole of O₂ occupies 22.4 liters of volume at S.T.P.

Therefore, the volume occupied by 0.0025 moles of O₂ is:

$= 0.0025 \times 22.4$

$=0.056l$

$=56 ml$

• The volume of O₂ gas evolved at STP is 56 ml or 0.056 l.