We need the reduction half-reaction of dichromate ion in acidic medium.

In acidic medium, $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ is reduced to $\mathrm{Cr}^{3+}$.

Start with: $\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+}$

Balance oxygen by adding water:

There are $7$ oxygen atoms on the left, so add $7\mathrm{H}_2\mathrm{O}$ on the right: $\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Balance hydrogen by adding $\mathrm{H}^+$:

Right side has $7\mathrm{H}_2\mathrm{O}$, i.e. $14$ H atoms, so add $14\mathrm{H}^+$ to the left: $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Balance charge by adding electrons:

• Left side charge $= -2 + 14 = +12$
• Right side charge $= 2 \times (+3) = +6$

To reduce left-side charge from $+12$ to $+6$, add $6e^-$ to the left: $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Comparing with $\mathrm{Cr}_2\mathrm{O}_7^{2-}+X\mathrm{H}^{+}+Ye^{\ominus} \rightarrow 2A+Z\mathrm{H}_2\mathrm{O}$ we get:

• $X = 14$
• $Y = 6$
• $Z = 7$
• $A = \mathrm{Cr}^{3+}$

Hence the correct option is: $\boxed{\text{B: } 14,6,7 \text{ and } \mathrm{Cr}^{3+}}$