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, and#Q# 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"# 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
#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...