Balancing RedOx Reactions in Acidic or Basic Aqueous Solution?
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!
Answer:
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)
Adding the spectacular ions
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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