1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Complete and balance the following reaction: Ag++AsH3⟶Ag↓+H3AsO3+H+

A
6Ag++AsH3+3H2O6Ag+6H++H3AsO3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6Ag++2AsH3+5H2O6Ag+10H++H3As2O3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6Ag++2AsH3+3H2O6Ag+6H++H3As2O3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 6Ag++AsH3+3H2O→6Ag↓+6H++H3AsO31. AsH3(g)→H3AsO3(aq)+(n)e− The number of electrons involved in this one is not so obvious, and we need to figure out the oxidation state of arsenic in AsH3 and H3AsO3 in order to deduce the number of electrons involved in this half-cell reaction. AsH3→As3− (each H is treated as H+ in this molecule) H3AsO3→As3+ (the anion is AsO3)3− This half-cell reaction can then be written as follows:As3−→As3++6e− 2. We can see that the silver half-cell needs to be multiplied by 6 to balance the number of electrons gained and lost: 6Ag+(aq)+6e−→6Ag(s) 3. We use this information on the two half-cells to start entering the balancing coefficients into the original equation, 6Ag+(aq)+AsH3(g)+H2O→6Ag(s)+H3AsO3(aq)+H+(aq) 4. The left side of the equation has a net charge of +6 and so, the right side must also have a net charge of +6. That means that the 6 Ag+ on the left side must be balanced by 6 H+ to make the net charge equal on both side. 6Ag+(aq)+AsH3(g)+H2O→6Ag(s)+H3AsO3(aq)+6H+(aq) 5. Now balance the oxygens and hydrogens by putting 3 in front of the H2O on the right side. 6Ag+(aq)+AsH3(g)+3H2O→6Ag(s)+H3AsO3(aq)+6H+(aq)Hence, the correct option is (A).

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Alkanes - Preparation
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program