JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 20)
The reaction of sulphur in alkaline medium is given below:
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The values of 'a' is _______. (Integer answer)
The values of 'a' is _______. (Integer answer)
Answer
12
Explanation
The two half reaction, one separately are as follows
$${S_8} + 16{e^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }}$$ (Reduction)
$${S_8} + 12{H_2}O\buildrel {} \over \longrightarrow 4{S_2}O_3^{2 - } + 24{H^ + } + 16{e^ - }$$
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$$2{S_8} + 12{H_2}O\buildrel {} \over \longrightarrow \mathop {8{S^{2 - }} + 4{S_2}O_3^{2 - }}\limits_{} + 24{H^ + }$$
For balancing in basic medium, Add an equal number of OH$$-$$ that of H+, we get
$$2{S_8} + 12{H_2}O + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 24{H_2}O$$
$$2{S_8} + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 12{H_2}O$$
or $${S_8} + 12O{H^ - }\buildrel {} \over \longrightarrow 4{S^2} + 2{S_2}O_3^{2 - } + 6{H_2}O$$ .... (i)
On comparing (i) with
$${S_8} + aO{H^ - }(aq)\buildrel {} \over \longrightarrow b{S^2}(aq) + c{S_2}O_3^{2 - } + d{H_2}O$$
We get, a = 12; b = 4; c = 2; d = 6
$${S_8} + 16{e^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }}$$ (Reduction)
$${S_8} + 12{H_2}O\buildrel {} \over \longrightarrow 4{S_2}O_3^{2 - } + 24{H^ + } + 16{e^ - }$$
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$$2{S_8} + 12{H_2}O\buildrel {} \over \longrightarrow \mathop {8{S^{2 - }} + 4{S_2}O_3^{2 - }}\limits_{} + 24{H^ + }$$
For balancing in basic medium, Add an equal number of OH$$-$$ that of H+, we get
$$2{S_8} + 12{H_2}O + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 24{H_2}O$$
$$2{S_8} + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 12{H_2}O$$
or $${S_8} + 12O{H^ - }\buildrel {} \over \longrightarrow 4{S^2} + 2{S_2}O_3^{2 - } + 6{H_2}O$$ .... (i)
On comparing (i) with
$${S_8} + aO{H^ - }(aq)\buildrel {} \over \longrightarrow b{S^2}(aq) + c{S_2}O_3^{2 - } + d{H_2}O$$
We get, a = 12; b = 4; c = 2; d = 6
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