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)"#

#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"#

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...