What would be the coefficient of H+ and H2O respectively in the balanced net ionic equations for the given redox reactions in acidic solution: S4O2−6(aq)+Al(s)→H2S(aq)+Al3+(aq)
A
20, 6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6, 20
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
10, 4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4, 10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 20, 6 Steps for Balancing redox reactions:
Identify the oxidation and reduction half .
Find the oxidising and reducing agent.
Find the n-factor of oxidising and reducing agent.
Balance atom undergoing oxidation and reduction.
Cross multiply the oxidising or reducing agent with n-factor.
Balance atoms other than oxygen and hydrogen.
Balancing oxygen atoms
Balancing hydrogen atoms
Balance charge
For acidic medium:
As soon as we add x H2O units, we add 2x H+ ions on the opposite side.
nf=(|O.S.Product−O.S.Reactant|×number of atom +2.5S4O2−6(aq)+0Al(s)→H2−2S(aq)++3Al3+(aq)
+2.5S4O2−6(aq)→H2−2S(aq) oxidation nf=(|−2−2.5|×4=18 0Al(aq)→+3Al3+(aq) reduction nf=(|3−0|×1=3 S4O2−6 is oxidising agent Al is reducing agent.
Balance atom undergoing oxidation and reduction. S4O2−6(aq)+Al(s)→4H2S(aq)+Al3+(aq)
Ratio of n- factor's =18:3
dividing the n-factors by common value in them. which is 3 in this case. ⇒Ratio of n- factor's =6:1
Cross mutiply the oxidising and reducing agents with ratio of n-factors. and balancing the main elements other than oxygen and hydrogen.
balance charge
charge in reactant side = -2+20=+18
charge in product side = +18
so the balanced equation is S4O2−6(aq)+6Al(s)+20H+→4H2S(aq)+6Al3+(aq)+6H2O(aq)
the coefficient of H+ and H2O respectively are 20 and 6.