Question

A current is passed through 500mL of an aqueous solution of $Ca{I}_{2}$. After sometime, it is observed that 50 millimoles of ${I}_{2}$ have been formed.
Which of the following statements is(are) correct?
A. The number of faradays of charge passed through the solution is 0.10 F.
B. The volume of dry ${H}_{2}$ at STP that has been formed during electrolysis is 1120 mL.
C. The pH of the solution is nearly 0.7.
D. The mass the calcium produced is 2.0g.

Answer 1

Hint: In this reaction, oxidation of ${I}^{-}$ takes place and the reduction of ${H}^{+}$ takes place. The products that are formed are iodine and hydrogen.

Complete step by step answer:
In this reaction, ${I}^{-}$ gets oxidized into ${I}_{2}$ by losing two electrons. Since it is an oxidation reaction it takes place at the anode. The reaction involved is shown below:
$ 2{ I }^{ - }\quad \longrightarrow \quad { I }_{ 2 }\quad +\quad 2{ e }^{ - }$

As oxidation is taking place, that means reduction will also take place. The reduction potential of hydrogen is more than that of calcium according to the electrochemical series. So, instead of calcium getting reduced, hydrogen gets reduced and is formed as a product. Since, it is a reduction reaction, it takes place at the cathode. The reaction involved is shown below:
$ 2{ H }^{ + }\quad +\quad 2{ e }^{ - }\quad \longrightarrow \quad { H }_{ 2 }$
Now, let us consider each point in the question given and which ones are true and why.
The number of faradays of charge passed through the solution is 0.10 F:
For the production 1 mole of ${I}_{2}$ the charge required is 2 F. Therefore, to produce 50 millimoles of ${I}_{2}$, the charge required or the faradays of charge passed through the solution is:
Charge required $= 2 \times 50 \times {10}^{-3}$
$\implies$ Charge required $= 0.10 F$
Therefore, the number of faradays of charge passed through the solution is 0.10 F. Thus, this statement is correct.
The volume of dry ${H}_{2}$ at STP that has been formed during electrolysis is 1120 mL:
Let us assume that 'm' millimole of hydrogen is getting deposited or produced.
$ m\quad =\quad \cfrac { 1 }{ 2 } \quad \times \quad no.\quad of\quad millimoles$
Substituting no. of milliF = 100 milliF
$\implies m\quad =\quad \cfrac { 1 }{ 2 } \quad \times \quad 100$
$\implies m = 50 millimoles$
Therefore, 50 millimoles of ${H}_{2}$ is produced.
Now, the volume of ${H}_{2} = no. of millimoles \times 22.4$
Substituting no. of millimoles = 50 millimoles
$\implies V = 50 \times 22.4$
$\implies V = 1120 mL$
Therefore, the volume of dry ${H}_{2}$ at STP that has been formed during electrolysis is 1120 mL. Thus, this statement is also correct.
The pH of the solution in nearly 0.7:
Now, ${Ca}^{2+}$ and ${OH}^{-}$ are left in the solution. We required 2 moles of ${H}^{+}$ ions to produce 1 mole of ${H}_{2}$. This means that 100 millimoles of ${H}^{+}$ are required to produce 50 millimoles of ${H}_{2}$.
Therefore, the concentration of ${OH}^{-}$ is given as:
$ { [OH] }^{ - }\quad =\quad \cfrac { 100 }{ 500 }$
$\implies [{OH}^{-}] = 0.2 M$
Now, the pOH is calculated as $pOH = -log[{OH}^{-}]$.
Substituting $[{OH}^{-}]$ = 0.2 M, we get
pOH = -log(0.2)
$\implies pOH = 0.7$
$\implies pH = 13.3$
Therefore, The pH of the solution is nearly 13.3. Thus, this statement is wrong.
The mass the calcium produced is 2.0g.
Since, calcium is not produced as a product in this reaction, therefore this statement is also wrong.
Therefore, options (a) and (b) are correct.

Note: While solving these types of questions, we need to keep in mind the reduction potentials of all the elements and also, how the oxidation and reduction of each element takes place. We also need to keep in mind which elements have the capability to get reduced and which elements have the capability to get oxidized.