# Problem: Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the strengths of the oxidizing and reducing agents:(1) I2(s) + 2e− ⟶ 2I−(aq)               E° = 0.53 V(2) S2O82−(aq) + 2e− ⟶ 2SO42−(aq)             E° = 2.01 V(3) Cr2O72−(aq) + 14H+(aq) + 6e− ⟶ 2Cr3+(aq) + 7H2O(l)           E° = 1.33 V

###### FREE Expert Solution

$\overline{){\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{cathode}}}{\mathbf{-}}{\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{anode}}}}$

E°cell > 0 → spontaneous in the forward direction
E°cell < 0 → nonspontaneous in the forward direction

Spontaneous: E°cathode > E°anode

S2O82−(aq) + 2e ⟶  2 SO42−(aq)            E° = 2.01 V ⟶  cathode
I2(s) + 2e ⟶ 2 I(aq)                               E° = 0.53 V ⟶ anode

E°cell = 1.48 V

S2O82−(aq) + 2e ⟶  2 SO42−(aq)             E° = 2.01 V ⟶  cathode
Cr2O72−(aq) + 14 H+(aq) + 6e ⟶ 2 Cr3+(aq) + 7 H2O(l)           E° = 1.33 V ⟶  anode

E°cell = 0.68 V

Cr2O72−(aq) + 14 H+(aq) + 6e ⟶ 2 Cr3+(aq) + 7 H2O(l)           E° = 1.33 V ⟶ cathode
I2(s) + 2e ⟶ 2 I(aq)                               E° = 0.53 V ⟶  anode

E°cell = 0.80 V

Cathode

Gain Electrons → Reduction → Oxidizing Agent

↑ E° → reduction → cathode

98% (377 ratings)
###### Problem Details

Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the strengths of the oxidizing and reducing agents:

(1) I2(s) + 2e ⟶ 2I(aq)               E° = 0.53 V

(2) S2O82−(aq) + 2e ⟶ 2SO42−(aq)             E° = 2.01 V

(3) Cr2O72−(aq) + 14H+(aq) + 6e ⟶ 2Cr3+(aq) + 7H2O(l)           E° = 1.33 V