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]
