ElectrochemistryHard
Question
Given:
H2O2 → O2 + 2H+ + 2e- E- = - 0.69 V
H2O2 + 2H+ + 2e- → 2H2O E- = - 1.77 V
I- → I2 + 2e-
Which of the following statements is(are) correct ?
H2O2 → O2 + 2H+ + 2e- E- = - 0.69 V
H2O2 + 2H+ + 2e- → 2H2O E- = - 1.77 V
I- → I2 + 2e-
Which of the following statements is(are) correct ?
Options
A.H2O2 behaves as an oxidant for I2/I-.
B.H2O2 behaves as an reductant for I2/I-.
C.I2/I- behaves as an reductant for H2O2.
D.None of these is correct.
Solution
Anode: 2I- → I2 + 2e-;
Anode: 2I- → I2 + 2e- ; E-a(oxid) = -0.535 ⇒ E-a(Red) = +0.535 V
Cathode: H2O2 + 2H+ + 2e- → 2H2O; E-c = 1.77 V
E-cell = (E-c - E-a)R = 1.77 - 0.53 = +ve
Note that:
Anode: H2O2 → O2 + 2H+ + 2e- ; E-a(oxid) = -0.69 V
Cathode: I2 + 2e- → 2I; E-c(red) = + 0.535 V
E-a(red) = + 0.69 V
E- = E-c - E-a = 0.535 - 0.69 = -ve
Anode: 2I- → I2 + 2e- ; E-a(oxid) = -0.535 ⇒ E-a(Red) = +0.535 V
Cathode: H2O2 + 2H+ + 2e- → 2H2O; E-c = 1.77 V
E-cell = (E-c - E-a)R = 1.77 - 0.53 = +ve
Note that:
Anode: H2O2 → O2 + 2H+ + 2e- ; E-a(oxid) = -0.69 V
Cathode: I2 + 2e- → 2I; E-c(red) = + 0.535 V
E-a(red) = + 0.69 V
E- = E-c - E-a = 0.535 - 0.69 = -ve
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