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)"
Reaction would be:
2 Answers
Explanation:
Based on voltaic charge table,
Well, I get
I'm assuming at
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
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)^@ = 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_(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
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, andQ is the reaction quotient.
Note that
First we get
Q = ((["Mg"^(2+)]//c^@)(P_(H_2)//P^@))/(["H"^(+)]//c^@)^2 (where we use
c^@ = "1 M" andP^@ = "1 atm" to makeQ unitless for use inlnQ .)
= (("0.20 M"//"1 M")("0.5 atm"//"1 atm"))/("0.300 M"//"1 M")^2
= 1.11
Now we proceed to calculate
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...