The standard reduction potentials at 298 K for the following half cells are given below :
Cr₂ O₇²⁻+14 H⁺+6 e⁻ → 2 Cr³⁺+7 H₂ O, E°=1.33 V
Fe³⁺(aq)+3 e⁻ → Fe, E°=-0.04 V; Ni²⁺(aq)+2 e⁻ → Ni, E°=-0.25 V; Ag⁺(aq)+e⁻ → Ag, E°=0.80 V; Au³⁺(aq)+3 e⁻ → Au, E°=1.40 V
Consider the given electrochemical reactions,
The number of metal(s) which will be oxidized be Cr₂ O₇²⁻, in aqueous solution is _________.

Solution

Chemistry ma’am · 2 steps

Dichromate can oxidise any metal whose reduction potential sits below its own 1.33 V.

1. The rule. In the cell, Cr₂O₇²⁻ is reduced at the cathode and the metal is oxidised at the anode. The reaction is spontaneous only when E°_(cell) is positive.

   E°_(cell) = E°_(Cr₂O₇²⁻/Cr³⁺) - E°ₘₑₜₐₗ = 1.33 V - E°ₘₑₜₐₗ

   Metal	E° of its ion (V)	E°_(cell) (V)	Oxidised?
   Fe	-0.04	1.33 + 0.04 = 1.37	yes
   Ni	-0.25	1.33 + 0.25 = 1.58	yes
   Ag	0.80	1.33 - 0.80 = 0.53	yes
   Au	1.40	1.33 - 1.40 = -0.07	no

2. So. Fe, Ni and Ag have positive E°_(cell). Au is a stronger oxidising couple than dichromate, so it is not oxidised. The answer is 3.

The common slip: 2. Counting only the metals with a negative E° (Fe and Ni) gives 2. Silver is oxidised too, because 0.80 V is still below 1.33 V.

The idea to keep. An oxidising agent attacks every species whose reduction potential is lower than its own. Compare the two E° values; their sign alone does not decide it.

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