Question

question 202 oxidation and reduction
the following chemical reaction was observed in a laboratory:
h₂s + cl₂ → s + 2 hcl
what is the reducing agent?
select one
cl₂
hcl
s
h₂s

Explanation:

Step1: Determine oxidation states

• In \(H_2S\), oxidation state of \(S\) is \(-2\) (since \(H\) is \(+1\) and \(2\times(+1)+x = 0\), so \(x=-2\)).
• In \(S\) (product), oxidation state of \(S\) is \(0\).
• In \(Cl_2\), oxidation state of \(Cl\) is \(0\).
• In \(HCl\), oxidation state of \(Cl\) is \(-1\) (since \(H\) is \(+1\) and \(+1 + y=0\), so \(y = -1\)).

Step2: Identify oxidation and reduction

• \(S\) in \(H_2S\) goes from \(-2\) to \(0\) (oxidation: loss of electrons).
• \(Cl\) in \(Cl_2\) goes from \(0\) to \(-1\) (reduction: gain of electrons).

Step3: Define reducing agent

• A reducing agent is a substance that gets oxidized (loses electrons) and causes the reduction of another substance.
• Since \(H_2S\) is oxidized ( \(S\) in \(H_2S\) loses electrons), \(H_2S\) is the reducing agent.

Answer:

\(H_2S\) (the option corresponding to \(H_2S\) in the multiple - choice list)