Electrode potentials (E°) are given below: Cu+/Cu=+0.52 V, Fe3+/Fe2+ = +0.77 V, (1/2)I2(s)/I- = 40.54 V, Ag+/Ag =+ 0.88 V. Based on the above potentials, strongest oxidizing agent will be :

Question

Electrode potentials (E°) are given below: Cu+/Cu=+0.52 V, Fe3+/Fe2+ = +0.77 V, (1/2)I2(s)/I- = 40.54 V, Ag+/Ag =+ 0.88 V. Based on the above potentials, strongest oxidizing agent will be : (A) Cu+ (B) Fe3+ (C) Ag3+ (D) I2

Tardigrade Admin · Accepted Answer

Higher the value of reduction potential stronger will be the oxidising hence based on the given values Ag+ will be strongest oxidizing agent.

Q. Electrode potentials (E $^{\circ}$ ) are given below :
$C u^{+} / C u = + 0.52 V,$
$F e^{3 +} / F e^{2 +} = + 0.77 V,$
$\frac{1}{2} I_{2} (s) / I^{-} = 40.54 V,$
$A g^{+} / A g = + 0.88 V .$
Based on the above potentials, strongest oxidizing agent will be :

$C u^{+}$

$F e^{3 +}$

$A g^{3 +}$

$I_{2}$

Higher the value of reduction potential stronger will be the oxidising hence based on the given values $A g^{+}$ will be strongest oxidizing agent.

Q. Electrode potentials (E∘) are given below : Cu+/Cu=+0.52V, Fe3+/Fe2+=+0.77V, 21​I2​(s)/I−=40.54V, Ag+/Ag=+0.88V. Based on the above potentials, strongest oxidizing agent will be :

Cu+

Fe3+

Ag3+

I2​