In acidic medium, K2Cr2O7 shows oxidising action as represented in the half reaction: Cr2O72- + XH+ + Ye 2A + ZH2O X, Y, Z and A are respectively :
- 14, 7, 6 and Cr3+
- 14, 6, 7 and Cr3+
- 8, 4, 6 and Cr2O3
- 8, 6, 4 and Cr2O3
Answer
(2) 14, 6, 7 and Cr3+
Solution
In acid, Cr2O72- (Cr = +6) is reduced to Cr3+ ⇒ A = Cr3+
Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O
O balance: 7 O on the left ⇒ Z = 7
H balance: 2 × 7 = 14 ⇒ X = 14
Charge balance: −2 + 14 − Y = +6 ⇒ Y = 6 (2 Cr × 3 e, +6 → +3)
Answer: (2)