Consider the following equations-2Fe2-H2O2- xA-yBin basic medium2MnO4-6H-5H2O2- x-C-y-D-z-Ein acidic mediumThe sum of the stoichiometric coefficients x- y- x- y- and z- for products A- B- C- D and E- respectively- is

[Fe^2+→ Fe^3++e^ ]× 2 H_2O_2+2e^ → 2OH^ 2Fe^2++H_2O_2→ 2Fe^3++ 2 OH^ x=2 y=2 [8H^++MnO_4^ +5e^ → Mn^2++4H_2O]× 2 [H_2O_2→ O_2(g)+2H^++2e^ ]× 5 ⇒ 16H^++ ...

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Byju's Answer

Standard XII

Chemistry

Van't Hoff Factor

Consider the ...

Question

Consider the following equations:
$2 F e^{2 +} + H_{2} O_{2} \to x A + y B$
(in basic medium)
$2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to x^{'} C + y^{'} D + z^{'} E$
(in acidic medium)

The sum of the stoichiometric coefficients $x, y, x^{'}, y^{'} and z^{'}$ for products $A, B, C, D and E, respectively$ , is

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Solution

$[F e^{2 +} \to F e^{3 +} + e^{-}] \times 2$

$\frac{H_{2} O_{2} + 2 e^{-} \to 2 H O^{Θ}}{2 F e^{2 +} + H_{2} O_{2} \to 2 F e^{3 +} + 2 H O_{(q ω)}^{Θ}}$

$x = 2 y = 2$

$[8 H^{+} + M n O_{4}^{-} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O] \times 2$

$[H_{2} O_{2} \to O_{2 (g)} + 2 H^{+} + 2 e^{-}] \times 5$

$\Rightarrow 16 H^{+} + 2 M n O_{4}^{-} + 5 H_{2} O_{2}$

$\to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2 (g)} + 10 H^{+}$

$\Rightarrow 6 H^{+} + 2 M n O_{4}^{-} + 5 H_{2} O_{2}$

$\to 2 M n^{2 +} + 8 H_{2} O + 50_{2 (g)}$

So $x^{'} = 2 y^{'} = 8 z^{'} = 5$

So $x + y + x^{'} + y^{'} + z^{'}$

$= 2 + 2 + 2 + 8 + 5$

$= 19$

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Consider the following equations: 2Fe2++H2O2→xA+yB (in basic medium) 2MnO−4+6H++5H2O2→x′C+y′D+z′E (in acidic medium) The sum of the stoichiometric coefficients x,y,x′,y′and z′ for products A,B,C,Dand E,respectively, is

[Fe2+→Fe3++e−]×2 H2O2+2e−→2HOΘ2Fe2++H2O2→2Fe3++2HOΘ(qω) x=2 y=2 [8H++MnO−4+5e−→Mn2++4H2O]×2 [H2O2→O2(g)+2H++2e−]×5 ⇒16H++2MnO−4+5H2O2 →2Mn2++8H2O+5O2(g)+10H+ ⇒6H++2MnO−4+5H2O2 →2Mn2++8H2O+502(g) So x′=2 y′=8 z′=5 So x+y+x′+y′+z′ =2+2+2+8+5 =19