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Representation of Cell

Balanced equa...

Question

Balanced equation for the reaction representing the reduction of nitrate ions to ammonium ions is:
$N O_{3}^{-} (a q) + x H^{+} (a q) + y e^{-} ⇌ N H_{4}^{+} (a q) + z H_{2} O$
What will be value of $x ?$

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Solution

For balancing the chemical reaction, first we have to balance the oxygen atom so, to balance the oxygen atom we have $z = 3$ , and reaction becomes
$N O_{3}^{-} + x H^{+} + y e^{-} \to N H_{4}^{+} + 3 H_{2} O$
now, in the product we hace total of 10 $H^{+}$ ions;
so, to balance the number of $H^{+}$ ion on left-hand side, the value of $x$ becomes 10.

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Similar questions

Q. In the reaction,

N H_{3 (a q)} + H_{2} O_{(l)} = N H_{4}^{+} (a q) + O H^{-} (a q)

the ammonium ions is the-

Q.
For the reduction of

N O_{3}^{-}

ion in an aqueous solution

E^{\circ}

is +0.96 V. Values of

E^{\circ}

for some metal ions are given below:

V^{2 +} (a q) + 2 e^{-} \to V; E^{0} = - 1.19 V

F e^{3 -} (a q) + 3 e^{-} \to F e; E^{\circ} = - 0.04 V

A u^{3 +} (a q) + 3 e^{-} \to A u; E^{\circ} = + 1.40 V

H g^{2 +} (a q) + 2 e^{-} \to H g; E^{0} = + 0.86 V

The pair(s) of metals that is/are oxidised by

N O_{3}^{-}

in aqueous solution is (are)

Q. For the reduction of

N O_{3}^{-}

ion in an aqueous solution,

E^{0} i s + 0.96 V .

E^{0}

for some metal ions are given below

V^{2 +} (a q) + 2 e^{-} \to V E^{0} = - 1.19 V

F e^{3 +} (a q) + 3 e^{-} \to F e E^{0} = - 0.04 V

A u^{3 +} (a q) + 3 e^{-} \to A u E^{0} = + 1.40 V

H g^{2 +} (a q) + 2 e^{-} \to H g E^{0} = + 0.86 V

The pair(s) metals that is(are) oxidized by

N O_{3}^{-}

in aqueous solution is(are):

Q. Which of the following metal cannot be oxidized by

N O_{3}^{-}

in aqueous solution?

For the reduction of

N O_{3}^{-}

ion in an aqueous solution,

E^{0}

+ 0.96 V

E^{0}

for metal ions are given below.

V^{2 +} (a q) + 2 e^{-} \to V (s); E^{0} = - 1.19 V

F e^{3 +} (a q) + 3 e^{-} \to F e (s); E^{0} = - 0.04 V

A u^{3 +} (a q) + 3 e^{-} \to A u (s); E^{0} = + 1.40 V

H g^{2 +} (a q) + 2 e^{-} \to H g (s); E^{0} = + 0.86 V

For the reduction of $N O_{3}^{⊖}$ ion in an aqueous solution, $E^{⊖}$ is +0.96 V, the values of $E^{⊖}$ for some metal ions are given below:
$(i) V^{2 +} (a q) + 2 e^{-} \to V; E^{⊖} = - 1.19 V$
$(i i) F e^{3 +} (a q) + 3 e^{-} \to F e; E^{⊖} = - 0.04 V$
$(i i i) A u^{3 +} (a q) + 3 e^{-} \to A u; E^{⊖} = + 1.40 V$
$(i v) H g^{2 +} (a q) + 2 e^{-} \to H g; E^{⊖} = + 0.86 V$

The pair(s) of metals that is/are oxidized by $N O_{3}^{⊖}$ in aqueous solution is/are
(IIT-JEE, 2010)

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