Question

# Balance the following redox reaction occurring in basic solution. I−(aq)+MnO−4(aq)→I2(aq)+MnO2(s) Express your answer as a chemical...

Balance the following redox reaction occurring in basic solution.
I−(aq)+MnO−4(aq)→I2(aq)+MnO2(s)

We are given redox reaction –

I-(aq) + MnO4- (aq) ------> I2(aq) + MnO2(s)

So first we need to determine the oxidation half reaction and reduction half reaction

Step 1) Assign the oxidation state to each element and write the two half reaction

I-(aq) + MnO4- (aq) ------> I2(aq) + MnO2(s)

Mn = +7                              Mn = +4

O = -2                                  O = -2

I = -1                                    I = 0

So Mn get reduced and I gets oxidized

So,

I-(aq) ------> I2(aq) ………oxidation half reaction

MnO4-(aq) ----> MnO2(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

2 I-(aq) ------> I2(aq)

MnO4-(aq) ----> MnO2(s)

Step 3) Balance the O by adding 1 H2O for 1 O

2 I-(aq) ------> I2(aq)

MnO4-(aq) ----> MnO2(s) + 2H2O(l)

Step 4) Balance the H by adding H+

2 I-(aq) ------> I2(aq)

MnO4-(aq) + 4H+(aq) ----> MnO2(s) + 2H2O(l)

Step 5) Balance the charge by adding electron

2 I-(aq) ------> I2(aq) + 2e-

MnO4-(aq) + 4H+(aq) + 3e- ----> MnO2(s) + 2H2O(l)

Step 6) Balance the electron in both half reaction – Multiply by reduction half reaction by 2 and oxidation half by 3

6 I-(aq) ------> 3 I2(aq) + 6e-

2 MnO4-(aq) + 8H+(aq) + 6e- ----> 2 MnO2(s) + 4H2O(l)

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6 I-(aq) + 2 MnO4-(aq) + 8H+ ----->3 I2(aq) + 2 MnO2(s) + 4H2O(l)

Now we need to added OH- from both side as the number of H+ present in the reaction

6 I-(aq) + 2 MnO4-(aq) + 8H+ + 8OH-(aq)----->3 I2(aq) + 2 MnO2(s) + 4H2O(l) + 8OH-(aq)

6 I-(aq) + 2 MnO4-(aq) +8H2O(l) ----->3 I2(aq) + 2 MnO2(s) + 4H2O(l) + 8OH-(aq)

6 I-(aq) + 2 MnO4-(aq) +4H2O(l) ----->3 I2(aq) + 2 MnO2(s) + 8OH-(aq)

So balanced redox reaction in basic medium is –

6 I-(aq) + 2 MnO4-(aq) +4H2O(l) ----->3 I2(aq) + 2 MnO2(s) + 8OH-(aq)

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