Acidified potassium dichromate oxidises hydrogen sulphide to sulphur. The balanced ionic equation for the reaction is:
\( \ce{Cr2O7^2-} + x\,\ce{H+} + 3\,\ce{H2S} \longrightarrow 2\,\ce{Cr^3+} + 3\,\ce{S} + y\,\ce{H2O} \)
What are the values of x and y?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → x = 8, y = 7
The full ionic equation is obtained by adding the half-reaction of dichromate to the half-reaction of the reducing agent, after making the electrons equal.
Cr: +6 in Cr2O72− → +3 in Cr3+; two Cr atoms gain 2 × 3 = 6 electrons.
S: −2 in H2S → 0 in S; three H2S molecules lose 3 × 2 = 6 electrons.
\( \displaystyle \ce{Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O} \)
\( \displaystyle \ce{3H2S -> 6H+ + 3S + 6e^-} \)
Adding the two, the 6H+ released by H2S cancel 6 of the 14H+, which leaves 8H+:
\( \displaystyle \ce{Cr2O7^2- + 8H+ + 3H2S -> 2Cr^3+ + 3S + 7H2O} \)
Check, oxygen: 7 = 7. Hydrogen: 8 + 6 = 14 = 2 × 7.
Check, charge: −2 + 8 = +6 on the left; 2 × (+3) = +6 on the right.
So x = 8 and y = 7.
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “…In acidic solution, its oxidising action can be represented as follows: Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O (Eo = 1.33V) Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III). The half-reactions are noted below:” … “3 H2S → 6H+ + 3S + 6e–”
“The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g., Cr2O72– + 14 H+ + 6 Fe2+ → 2 Cr3+ + 6 Fe3+ + 7 H2O”