Question

\[ S_2O_3^{2-}(aq) + OH^{-}(aq) \rightarrow SO_4^{2-}(aq) + H_2O(l) + e^{-} \]  
After the above half reaction is balanced, which of the following are the coefficients of \(OH^{-}\) and \(SO_4^{2-}\) respectively?

• A. \(8,\;3\)
• B. \(6,\;2\)
• C. \(10,\;2\)
• D. \(5,\;2\) \bigskip

Hint:

In basic medium, balance oxygen with \(H_2O\), hydrogen with \(OH^{-}\), and finally balance charge using electrons.

Answer 1

The Correct Option is C

Concept: For balancing redox reactions in basic medium:
• Balance atoms other than O and H.
• Balance oxygen using \(H_2O\).
• Balance hydrogen using \(OH^-\).
• Balance charge using electrons.

Step 1: Balance sulphur atoms.
There are two sulphur atoms in \[ S_2O_3^{2-} \] Hence, \[ S_2O_3^{2-} \rightarrow 2SO_4^{2-} \]

Step 2: Balance oxygen atoms.
Left side oxygen atoms: \[ 3 \] Right side oxygen atoms: \[ 2\times4=8 \] Add \(5H_2O\) to left side: \[ S_2O_3^{2-}+5H_2O \rightarrow 2SO_4^{2-} \]

Step 3: Balance hydrogen in basic medium.
Left side contains \[ 10H \] Add \[ 10OH^- \] to the right side. \[ S_2O_3^{2-}+5H_2O \rightarrow 2SO_4^{2-}+10OH^- \]

Step 4: Balance charge using electrons.
Left side charge: \[ -2 \] Right side charge: \[ 2(-2)+10(-1)=-14 \] Difference: \[ 12 \] Add \(12e^-\) on right side. Balanced half reaction: \[ S_2O_3^{2-}+10OH^- \rightarrow 2SO_4^{2-}+5H_2O+8e^- \] Thus coefficients required are \[ OH^- =10 \] and \[ SO_4^{2-}=2 \]

Step 5: Write the final answer.
Hence, \[ \boxed{(10,\;2)} \] Therefore the correct option is \[ \boxed{\text{(C)}} \]