ElectrochemistryEasy
Question
In the balanced chemical reaction, IO3- + aI- + bH+ → cH2O + dI2 a, b, c and d respectively to
Options
A.5, 6, 3, 3
B.5, 2, 6, 3
C.3, 5, 3, 6
D.5, 6, 5, 5
Solution
+5IO3- + I- + H+ → H2O + 0I2
(A) Oxidation half cell
I- → I2
(i) Balancing the number of atoms
2I- → I2
(ii) Balancing charge
2I- → I2 + 2e-
(B) Reduction half reaction
IO3- + H+ → H2O + I2
(i) Balancing number of charge
2IO3- + 12H+ → 6H2O + I2
(ii) Balancing the charge
2IO3- + 12H+ + 10e- → 6H2O + I2
Multiplying the balanced oxidation half reaction by 5 and adding it to balanced reduction half reaction
2I- → I2 + 2e- × 5
2IO3- + 12H+ + 10e- → 6H2O + I2
2IO3- + 10I- + 12H+ → 6I2 + 6H2O
or IO3- + 5I- + 6H+ → 3I2 + 3H2O
a = 5, b = 6, c = 3 and d = 3
(A) Oxidation half cell
I- → I2
(i) Balancing the number of atoms
2I- → I2
(ii) Balancing charge
2I- → I2 + 2e-
(B) Reduction half reaction
IO3- + H+ → H2O + I2
(i) Balancing number of charge
2IO3- + 12H+ → 6H2O + I2
(ii) Balancing the charge
2IO3- + 12H+ + 10e- → 6H2O + I2
Multiplying the balanced oxidation half reaction by 5 and adding it to balanced reduction half reaction
2I- → I2 + 2e- × 5
2IO3- + 12H+ + 10e- → 6H2O + I2
2IO3- + 10I- + 12H+ → 6I2 + 6H2O
or IO3- + 5I- + 6H+ → 3I2 + 3H2O
a = 5, b = 6, c = 3 and d = 3
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