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Question: What is the oxidation state of tin \(\left( {Sn} \right)\) when current of \(2A\) is passed through ...

What is the oxidation state of tin (Sn)\left( {Sn} \right) when current of 2A2A is passed through molten tin salt for 1.93×104s1.93 \times {10^4}s ? Atomic weight of the tin is 119119.

Explanation

Solution

Michael Faraday gave two laws of electrolysis. From these laws, we can calculate the oxidation state of tin. Faraday’s first law states that the mass of the chemical substance deposited is directly proportional to the current. Faraday’s second law states that the amount of substances deposited by the passage of the same amount of electric current will be proportional to their equivalent weights. Combining the two laws we can calculate the oxidation state of SnSn .

Formula Used:
We will use the formula,
WEq=I×t96500\dfrac{W}{{Eq}} = \dfrac{{I \times t}}{{96500}}
Eq=A.W.nEq = \dfrac{{A.W.}}{n}

Complete answer: We have studied in electrochemistry that Michael Faraday has given certain laws that are known as laws of Electrolysis. Electrolysis is a process in which a direct current is passed through a substance that is ionic in nature. It can be in a molten state or dissolved in a suitable solvent. When a direct current is passed through the molten state, it results in the separation of ions and chemical reaction at respective electrodes. Michael Faraday gave two laws of Electrolysis.

Using Faraday’s laws, we get
WEq=I×t96500\dfrac{W}{{Eq}} = \dfrac{{I \times t}}{{96500}}……(1)\left( 1 \right)
W=W = Weight of deposited tin
Eq=Eq = The equivalent weight of tin
I=I = Current
t=t = Time
96500=96500 = Faraday’s Constant
Faraday’s Constant is equal to the charge carried by 11 mol of electrons. Its SISI unit is Cmol1Cmo{l^{ - 1}} .
In the question, we are given I=2AI = 2A , t=1.93×104st = 1.93 \times {10^4}s and W=23.8gW = 23.8g .
Putting these values in (1)\left( 1 \right)
WEq=I×t96500\dfrac{W}{{Eq}} = \dfrac{{I \times t}}{{96500}}

& \Rightarrow \dfrac{{23.8}}{{Eq}} = \dfrac{{2 \times 1.93 \times {{10}^4}}}{{96500}} \cr & \Rightarrow Eq = \dfrac{{23.8 \times 96500}}{{2 \times 1.93 \times {{10}^4}}} \cr} $$ From here we get Equivalent weight that is $59.5g$. Equivalent weight is given by the formula $Eq = \dfrac{{A.W.}}{n}$……$\left( 2 \right)$ $A.W. = $ The atomic weight of tin $n = $ Oxidation of tin In the question atomic weight of tin given $ = 119$, using the equation $\left( 2 \right)$ $\eqalign{ & \Rightarrow 59.5 = \dfrac{{119}}{n} \cr & \Rightarrow n = 2 \cr} $ Thus oxidation state of $Sn$ is $2$ **Note:** According to faraday’s first law, the mass of the chemical substance deposited is directly proportional to the current. When we remove proportionality, we get $Z$ a proportionality constant. Here $Z = \dfrac{{Eq}}{F}$ $Eq = $ Equivalent weight $F = $Faraday’s constant