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Consider the reaction mg(s)+fe2+(aq)→mg2+(aq)+fe(s) at 79 ∘c , where [fe2+]= 3.50 m and [mg2+]= 0.310 m . part a what is the value for the reaction quotient, q, for the cell? express your answer numerically.

Respuesta :

arthurpdc
The following reaction is given:

[tex]Mg_{(s)}+Fe^{2+}_{(aq)}\to Mg^{2+}_{(aq)}+Fe_{(s)}[/tex]

So, the reaction quotient is:

[tex]Q=\dfrac{[Products]}{[Reactants]}\\\\ Q=\dfrac{[Mg^{2+}]}{[Fe^{2+}]}[/tex]

Note that in the reaction quotient formula we only put the substances in the aqueous state. The substances in the solid state are not considered.

Then, we'll use the values of the statement:

[tex]Q=\dfrac{[Mg^{2+}]}{[Fe^{2+}]}=\dfrac{0.31~M}{3.50~M}\\\\ Q\approx0,08857\\\\ \boxed{Q\approx8,857\times10^{-2}}[/tex]
dsdrajlin

For the reaction Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s),

when [Fe²⁺]= 3.50 M and [Mg²⁺]= 0.310 M, the reaction quotient is 0.0886.

Let's consider the following balanced equation.

Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)

The reaction quotient (Q) is the ratio of the molar concentrations of the product species over those of the reactant species involved in the chemical reaction, taking stoichiometric coefficients of the reaction into account as exponents of the concentrations. It only includes gases and aqueous species.

Given [Fe²⁺]= 3.50 M and [Mg²⁺]= 0.310 M, the reaction quotient for this cell is:

[tex]Q = \frac{[Mg^{2+} ]}{[Fe^{2+} ]} = \frac{0.310}{3.50} = 0.0886[/tex]

For the reaction Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s),

when [Fe²⁺]= 3.50 M and [Mg²⁺]= 0.310 M, the reaction quotient is 0.0886.

Learn more: https://brainly.com/question/5144419

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