# Balancing RedOx Reactions in Acidic or Basic Aqueous Solution?

Compose the balanced reaction equation for the following reaction occurring in ACIDIC aqueous medium.
Reduce all coefficients to the lowest possible integral values.
Include H2O(l) and H+(aq) appropriately to complete the equation.

Fe(s) + Cl−(aq) → HFeCl4(aq) + H2(g)

Compose the balanced reaction equation for the following reaction occurring in BASIC aqueous medium.
Reduce all coefficients to the lowest possible integral values.
Include H2O(l) and OH−(aq) appropriately to complete the equation.

Fe(CN)64−(aq) + Ce4+(aq) → Ce(OH)3(s) + Fe(OH)3(s) + CO32−(aq) + NO3−(aq)

Any help is appreciated!
Thanks!

1)
Here the reaction is
Fe(s) + Cl(-) -----> HFeCl4 + H2(g)

All the solutions are aqueous. So, on adding a dummy H(+) ion on LHS, we have
Fe + Cl(-) + H(+) -----> HFeCl4 + H2

Here, identify the species that are getting oxidised and reduced. They are Fe and H respectively.
H is also being partially oxidised. Only a part of H is appearing in the reduced form (H2) and the other part is in HFeCl4. This H in HFeCl4 must be in 1:1 ratio with Fe

Let x moles of Fe be oxidised

Writing down the oxidation reaction
xFe ----> xFe(+3) + 3x e(-)

The reduction reaction is
(x+y) H(+) + y e(-) ------> x H(+) + y H(0)

To balance, we have 3x = y (The number of electrons in reduction and oxidation reaction is the same)

The reduction reaction can be re-written as
4x H(+) + 3x e(-) ----------> x H(+) + 3x H(0)
Adding to the oxidation reaction we have
Fe + 4H(+) -----> H(+) + 3H(0) + Fe(+3)

Rewriting the spectacular ions, we have
Fe + 4H(+) + Cl(-) ----> HFeCl4 + 1.5 H2

2Fe + 8H(+) + 8Cl(-) -----> 2HFeCl4 + 3H2

2)
Again find the species that are getting oxidised. They are C and N
The oxidation reactions can be written as
C(+2) -----> C(+4) + 2e(-)
N(-3) ------> N(+5) + 8e(-)
As these 2 are in the ratio 1:1, just add up these 2 equations
C(+2) + N(-3) ------> C(+4) + N(+5) + 10e(-) .(I)

Also, we have Fe being oxidised from the +2 state to the +3 state.
Fe(+2) ----> Fe(+3) + e(-) ..(II)

But the ratio of Fe:CN is 1:6. So, multiplying equation number (I) by 6 and adding to II we have

6 C(+2) + 6N(-3) + Fe(+2) -----> 6C(+4) + 6N(+5) + Fe(+3) + 61 e(-)

The only species reduced is Ce, from (+4) to (+3)
Ce(+4) + e(-) -----> Ce(+3)
Multiplying this by 61 and adding to the previous equation we have

61Ce(+4) + [Fe(CN)6 (-4)] ----> 61Ce(+3) + 6CO3(-2) + 6NO3(-) + Fe(+3)

61Ce(+4) + [Fe(CN)6 (-4)] ------> 61 Ce(OH)3 + Fe(OH)3 + CO3(-2) + NO3(-)

Balance the H and O by adding H(+) and H2O

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