Based on the data given below :
E_(Cr₂ O₇²⁻ / Cr³⁺)°=1.33 V, E_(Cl₂ / Cl⁽⁻⁾)°=1.36 V; E_(MnO₄⁻ / Mn²⁺)⁰=1.51 V, E_(Cr³⁺ / Cr)°=-0.74 V
the strongest reducing agent is :

Solution

Chemistry ma’am · 3 steps

A reducing agent sits on the right of a couple; the most negative E° gives the strongest one.

1. Where reducing agents sit. Each E° is written for a reduction: oxidised form + electrons gives reduced form. A reducing agent gives electrons away, so it must be the reduced form, the species on the right of each couple.
2. Read the sign. A low or negative E° means the couple does not want to be reduced, so its reduced form readily gets oxidised. The more negative the E°, the stronger the reduced form is as a reducing agent.

   Couple	E° (V)	Reduced form (reducing agent)
   MnO₄⁻/Mn²⁺	+1.51	Mn²⁺, weakest
   Cl₂/Cl⁻	+1.36	Cl⁻
   Cr₂O₇²⁻/Cr³⁺	+1.33	Cr³⁺
   Cr³⁺/Cr	−0.74	Cr, strongest

3. So. Cr³⁺/Cr has the only negative E°, so its reduced form Cr is the strongest reducing agent. The answer is (c).

The idea to keep. Strongest reducing agent: lowest reduction potential, then take the reduced form. Strongest oxidising agent: highest reduction potential, then take the oxidised form.

Answer from the NTA final answer key. Solution by PrepWiser.