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A certain metal M forms a soluble sulfate salt M2SO4. Suppose the left half cell of a galvanic cell apparatus is filled with a 50.0 mM solution of M2SO4 and the right half cell with a 5.00 M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them The temperature of the apparatus is held constant at 20.0 °C.
Which electrode will be positive?
What voltage will the voltmeter show?

Sagot :

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Answer:

Explanation:

For a certain metal M by which the electrodes and the solution M_2SO_4 the cell notation is:

                                    SALT BRIDGE

     ANODE                         ↓                    CATHODE

[tex]M(s)|M_2SO_4_{(aq)} (50.0 \ mM)\Big {|} \Big {|} M_2SO_4_{(aq)} (5.00 \ M ) |M(s)[/tex]

Since the standard electrode potential is less positive on the left side, a negative anode electrode is used, and oxidation occurs at the anode.

[tex]i.e \ \ M(s) \to M^{2+} _{(aq)} + 2 e^- \ \ \ (oxidation)[/tex]

The right side of the cell cathode is placed which is positive in sign and aids in reduction since the normal reduction potential is less negative.

[tex]i.e \ \ M^{2+} _{(aq)} + 2 e^- \to M(s) \ \ \ (reduction)[/tex]

As a result, the correct answer for the positive electrode is the right side.

[tex]M_2SO_4 \to M^{2+} + SO_4^{2-}[/tex]

Using Nernst Equation:

[tex]E_{cell} = - \dfrac{RT}{nF }log ( \dfrac{reduction \ half }{oxidation \ half})[/tex]

[tex]E_{cell} = - \dfrac{2.303 \times 8.314 \times 293}{2 \times 96485 }log ( \dfrac{5.0 0 }{50 \times 10^{-3}})[/tex]

[tex]E_{cell} = -0.058 \ V[/tex]

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