What is E_"cell" for the following battery? Mg"(s)"|Mg^{2+}(0.20"M")||H_2(0.5"atm")" "H^{+}(0.300"M")|Pt"(s)"

Mg"(s)"|Mg^{2+}(0.20"M")||H_2(0.5"atm")" "H^{+}(0.300"M")|Pt"(s)"

Reaction would be:
Mg"(s)"+2H^{+}\toMg^{2+}"(aq)"+H_2"(g)"

2 Answers
Jul 26, 2018

2.36" V"

Explanation:

E_"cell"=E°_"cell"-(0.0592" V")/n\logQ, and n=2 (2 e^{-})

Based on voltaic charge table,
E°_"cell"=2.356+0.00=2.356"V" (half-reactions of magnesium and hydrogen, added together)

\log(Q)=\log(([Mg^(2+)]P_(H_2))/[H^(+)]^2)=\log(((0.20)(0.5"atm"))/(0.300)^2)\approx0.045758


\thereforeE_"cell"=2.356-0.0592/2(0.045758)\approx2.356-0.00135"V"\approx2.36" V"

Jul 30, 2018

Well, I get E_(cell) ~~ E_(cell)^@, because Q ~~ 1.


I'm assuming at "298.15 K" and "1 atm".

In America, we write cell notation to have electrons flow from left to right. Thus, the left-hand side is the anode and right-hand side is the cathode, and we wish for "Mg"(s) to get oxidized by "H"^(+).

So, the reaction is:

2"H"^(+)(aq) + 2e^(-) -> "H"_2(g), E_(red)^@ = "0.00 V"
ul("Mg"(s) -> "Mg"^(2+)(aq) + 2e^(-)), E_(red)^@ = -"2.37 V"
2"H"^(+)(aq) + "Mg"(s) -> "Mg"^(2+)(aq) + "H"_2(g)

My preferred way to get E_(cell)^@ is:

E_(cell)^@ = E_(red)^@ + E_(o x)^@

= "0.00 V" + [-(-"2.37 V")]

= +"2.37 V"

Alternatively, a no-brainer way would be to subtract the LESS positive (MORE negative) E_(red)^@ from the MORE positive (LESS negative) E_(red)^@ to guarantee the spontaneous reaction.

E_(cell)^@ = E_("cathode")^@ - E_"anode"^@

= "0.00 V" - (-"2.37 V")

= +"2.37 V"

Goodie, that was the easy part. Now since we want nonstandard, nonequilibrium conditions at "298.15 K", we use the Nernst equation.

E_(cell) = E_(cell)^@ - (RT)/(nF)lnQ

where R is the universal gas constant, T is temperature in "K", n is the mols of electrons PER mol of atom, F = "96485 C/mol e"^(-) is Faraday's constant, and Q is the reaction quotient.

Note that ln x ~~ 2.303logx.

First we get Q.

Q = ((["Mg"^(2+)]//c^@)(P_(H_2)//P^@))/(["H"^(+)]//c^@)^2

(where we use c^@ = "1 M" and P^@ = "1 atm" to make Q unitless for use in lnQ.)

= (("0.20 M"//"1 M")("0.5 atm"//"1 atm"))/("0.300 M"//"1 M")^2

= 1.11

Now we proceed to calculate E_(cell).

color(blue)(E_(cell)) = "2.37 V" - ("8.314 V"cdotcancel"C""/"cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/((2 cancel("mol e"^(-)))/(cancel"1 mol Mg") cdot 96485 cancel"C""/"cancel("mol e"^(-))) ln (1.11)

= "2.37 V" - "0.0257 V"/("2 e"^(-)) ln (1.11)

= "2.37 V" - cancel("0.0592 V"/("2 e"^(-)) log(1.11))^"small"

= color(blue)"2.37 V" again...