In acidic medium, \(K _2 Cr _2 O _7\) shows oxidising action as represented in the half reaction \[Cr _2 O _7{ }^{2-}+ X…

Given half-reaction in acidic medium:

Cr₂O₇²⁻ + XH⁺ + Ye⁻ → 2A + ZH₂O

We know standard reduction of dichromate in acidic medium:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Now comparing,

X = number of H⁺ = 14
Y = number of electrons = 6
Z = number of H₂O = 7
A = product formed = Cr³⁺

So values are:
X = 14, Y = 6, Z = 7, A = Cr³⁺