Question

In the following reaction for the preparation of gold sol, \(a,\ b,\ c,\ x,\ y,\) and \(z\) are \[ aAuCl_3+bHCHO+cH_2O\rightarrow xAu(sol)+yHCO_2H+zHCl \]

• A. \(a=2,\ b=3,\ c=3,\ x=2,\ y=3,\ z=6\)
• B. \(a=2,\ b=3,\ c=2,\ x=2,\ y=3,\ z=4\)
• C. \(a=2,\ b=2,\ c=2,\ x=2,\ y=2,\ z=4\)
• D. \(a=3,\ b=2,\ c=2,\ x=3,\ y=2,\ z=6\)

Hint:

In redox reactions used for sol preparation, balance the metal and halogen atoms first, then balance carbon, oxygen, and hydrogen atoms.

Answer 1

The Correct Option is A

Step 1: Write the unbalanced equation and identify redox changes.
The reaction is $aAuCl_3 + bHCHO + cH_2O \rightarrow xAu\ (\text{sol}) + yHCO_2H + zHCl$. Gold is reduced from $+3$ to $0$, and formaldehyde (HCHO) is oxidized to formic acid ($HCO_2H$).
Step 2: Balance gold atoms.
Set $a = 2$, so $x = 2$.
Step 3: Balance chlorine atoms.
With $a = 2$, there are $2 \times 3 = 6$ chlorine atoms on the left, so $z = 6$.
Step 4: Balance electrons to find $b$ and $y$.
Each Au goes from $+3$ to $0$: gain of 3 electrons. For 2 Au atoms: total gain = 6 electrons. Each HCHO is oxidized to $HCO_2H$ (C goes from $0$ to $+2$): loss of 2 electrons per molecule. To balance: $6 / 2 = 3$ molecules of HCHO. So $b = 3$. By carbon balance: $y = 3$.
Step 5: Balance oxygen to find $c$.
Oxygen on the right: $3HCO_2H = 3 \times 2 = 6$ O atoms. Oxygen from $3HCHO = 3$ O atoms. Oxygen from water: $6 - 3 = 3$. So $c = 3$.
Step 6: Verify hydrogen balance.
Left: $3HCHO$ gives 6 H, $3H_2O$ gives 6 H. Total = 12 H. Right: $3HCO_2H$ gives 6 H, $6HCl$ gives 6 H. Total = 12 H. Balanced.
Step 7: State the final answer.
Balanced: $2AuCl_3 + 3HCHO + 3H_2O \rightarrow 2Au + 3HCO_2H + 6HCl$. \[ \boxed{a=2,\ b=3,\ c=3,\ x=2,\ y=3,\ z=6} \]