Question

A current of $1.40ampere$ is passed through $500ml$ of $0.180M$ solution of zinc sulphate for $200seconds$. What will be the molarity of $Z{n}^{2+}$ions after the deposition of zinc?

• A. $0.154M$
• B. $0.177M$
• C. $2M$
• D. $0.180M$

Answer 1

The correct option is B

$0.177M$

Step 1: Given data

Electric current flowing through zinc sulphate solution, $i=1.40A$

Time duration during which current flows through zinc sulphate solution, $t=200s$

Volume of the zinc sulphate solution, $v=500ml=0.5l$

Molarity of the zinc sulphate solution, $M_{zs}=0.180M$

Step 2: Assumptions

Equivalent weight$=w_e$

Molecular weight$=m$

Valency$=N$

Number of moles of zinc deposited $=n_{zd}$

Initial moles of the ($Z{n}^{2+}$) present in the solution $=n_z$

Number of moles of zinc left$=n_{zL}$

Molarity of ($Z{n}^{2+}$) ions after the deposition of zinc, $M_{zd}=?$

Step 3: Calculation of the molarity of $Z{n}^{2+}$ions after deposition of zinc

The given electrochemical equation can be written as

$Z{n}^{2+}+2\stackrel{-}{e}\to Zn$

From the above reaction, it is clear that the valency, $N=2$ ………….(s)

The weight ($w$) of the zinc deposited is given by the basic equation of “Faraday's first law of electrolysis”.

$w=z\times i\times t$ ……………………….(a)

Where Z is the proportionality constant as is called as “electrochemical equivalent” and is related to the equivalent weight by the equation

$z=\frac{w_e}{96500}$………………………..(b)

Equivalent weight is related to the molecular weight and the valency by the equation

$w_e=\frac{m}{N}$ …………………………….(c)

Using equation (c) in the equation (b), we get

$z=\frac{m}{N\times 96500}$ …………………(d)

Substituting equation (d) in equation (a), we get

$w=\frac{m}{N\times 96500}\times i\times t$ …………………………………(e)

Substituting the equation (s) and given values in equation (e), we get

$w=\frac{m}{2\times 96500}\times 1.40A\times 200s$

$$\begin{gathered} \frac{w}{m}=\frac{1.40\times 200}{2\times 96500} \\ \frac{w}{m}=0.00145mol \end{gathered}$$

Number of moles of zinc deposited, $n_{zd}=0.00145mol$ …………………….(f)

Initial moles of ($Z{n}^{2+}$) present in the solution will be equal to the product of the molarity of the zinc sulphate solution and its volume

$n_z=M_{zs}\times v$

Substituting the given values in the above equation, we get

$n_z=0.18\times 0.5$

$n_z=0.09mol$…………..(g)

The number of moles of zinc left will be equal to the difference between the initial moles of ($Z{n}^{2+}$) ions present in the solution and the number of moles of the zinc deposited

$n_{zL}=n_z-n_{zd}$ ………………….(h)

Using equations (f) and (g) in equation (h), we get

$$\begin{gathered} n_{zL}=0.09mol-0.00145mol \\ {n}_{zl}=0.08855mol \end{gathered}$$

The molarity of ($Z{n}^{2+}$) ions after the deposition of zinc will be equal to the ratio of the number of moles of zinc left and the volume of the zinc sulphate solution.

$M_{zd}=\frac{n_{zl}}{v}$

Substituting the given and obtained values in the above equation, we get

$$\begin{gathered} M_{zd}=\frac{0.008855mol}{0.5lt} \\ {M}_{zd}=0.177M \end{gathered}$$

Hence, option (B) is correct.