Question

under the following conditions is the reaction more spontaneous of le

20pts (4) · Balance the following redox reactions. Show all work in all steps. (a) write the two half reactions. (b) identify which is the oxidation half reaction and which is the reduction half reaction. (c) show the number of electrons involved in each half reaction. (d) write the overall (net) balanced reaction ***(aq) + H2QmO4(s) → X4+ (aq) + Qm(s) in basic soln. MO2 (aq) + 23+ (aq) → MO3(aq) + Z2+ (aq) in acid soln

Answer 1

Answer:-

As we know that in ion-electron method, we balanced chemical equations in two different ways.

(I) - Acidic medium :- in which whole chemical reaction is balanced by the H+ and H2O.

(II)- Basic medium :- in which whole chemical reaction is balanced by the OH- and H2O.

For balancing the chemical reaction , first we splits chemical reaction into two half reaction. Thereafter first we balance the atoms in both sides of half reaction and then balance the charge. After balancing both half reaction we add both half reaction to get net balanced reaction.

(1) -

Given:-

X3+(aq) + H2QmO4(s) $\rightarrow$   X4+(aq) + Qm(s) ; (basic solution)

First oxidation half reaction :-

X3+(aq) $\rightarrow$ X4+(aq) + e- ----------------------------------------(1)

Second reduction half reaction :-

H2QmO4(s) + 2H2O(l) + 6e-   $\rightarrow$    Qm(s) + 6OH-(aq) ---------------------------------------------(2)

from equation no. 1 and 2

(  X3+(aq) $\rightarrow$ X4+(aq) + e- ) $\times$ 6

( H2QmO4(s) + 2H2O(l) + 6e-   $\rightarrow$    Qm(s) + 6OH-(aq) ) $\times$ 1

we get

6X3+(aq) $\rightarrow$ 6X4+(aq) + 6e-

H2QmO4(s) + 2H2O(l) + 6e-   $\rightarrow$    Qm(s) + 6OH-(aq)

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6X3+(aq) + H2QmO4(s) + 2H2O(l)   $\rightarrow$ 6X4+(aq) + Qm(s) + 6OH-(aq) (Overall net balanced reaction)

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total no. of electrons involved in each reaction (n) = 6

(2) -

Given:-

MO2-(aq) + Z3+(aq) $\rightarrow$   MO3-(aq) + Z2+(aq) ; (acidic solution)

First oxidation half reaction :-

MO2-(aq)   + H2O(l)   $\rightarrow$ MO3-(aq)  + 2H+(aq)   + 2e- ----------------------------------------(1)

Second reduction half reaction :-

Z3+(aq) + e-   $\rightarrow$    Z2+(aq) ---------------------------------------------------------(2)

from equation no. 1 and 2

(MO2-(aq)   + H2O(l)   $\rightarrow$ MO3-(aq)  + 2H+(aq)   + 2e- ) $\times$ 1

( Z3+(aq) + e-   $\rightarrow$    Z2+(aq) ) $\times$ 2

we get

MO2-(aq)   + H2O(l)   $\rightarrow$ MO3-(aq)  + 2H+(aq)   + 2e-

2Z3+(aq) + 2e-   $\rightarrow$ 2Z2+(aq)  

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MO2-(aq) + 2Z3+(aq) + H2O(l)   $\rightarrow$ MO3-(aq) + 2Z2+(aq) + 2H+(aq) (Overall net balanced reaction)

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total no. of electrons involved in each reaction (n) = 2