When the following equation is balanced: Cr⁺³ + ClO⁻¹ → CrO₄⁻² + Cl⁻¹ What is the correct coefficient in front of Cl⁻¹?

CrO₄²⁻ is stable in basic media. Hence, the reaction medium is basic.

oxidation half equation: ClO⁻ + H₂O + 2 e⁻ → Cl⁻ + 2 OH⁻ …… [1]
Reduction: half equation: Cr³⁺ + 8 OH⁻ → CrO₄²⁻ + 4 H₂O + 3 e⁻ …… [2]

[1]×3 + [2]×2, and cancel 6 e⁻, 3 H₂O and 6 OH⁻ on the both sides. Then, the balanced equation is:
2 Cr³⁺ + 3 ClO⁻ + 10 OH⁻ → 2 CrO₄²⁻ + 3 Cl⁻ + 5H₂O

The coefficient in front of Cl⁻ = 3

I think it's 4

You have 1 Cr on the left and 1 on the right= balanced
You have 1 Cl on the left and 1 on the right = balanced
You have 1 O on the left and 4 on the right = not balanced

Need 4 Oxygens on the left = xClO by 4

Now you have 4 Cl on the left and 1 on the right

xCl by 4 on the right

= Cr + 4ClO -> CrO4 + 4Cl

(Include your other numbers I couldn't put those in)