Question

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)"`

Answer 1

`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"`

Answer 2

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...