# What is the value of #E_(cell)# at #"25"^oC# after 5.0 hours of operation at 3.0 A...?

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"The std potential (#E_(cell)^o# ) of a voltaic cell based on the #Zn# /#Cu^(2+)# ion reaction

#Zn(s)+Cu^(2+)(aq)\toZn^(2+)(aq)+Cu(s)#

is #"1.10 V"# at #25^@ "C"# .

What is the value of #E_(cell)# at #"25"^oC# after 5.0 hours of operation at 3.0 A, if #[Cu^(2+)]_0# is 1.90 M and #[Zn^(2+)]_0# is 0.100 M? Assume the volume in both the anode and the cathode compartments is a constant 1.00 L (each)."

See below answer (from me) for my work.

"The std potential (

is

What is the value of

See below answer (from me) for my work.

##### 2 Answers

#### Explanation:

**My work:**

Currently, I have calculated via dimensional analysis, and after 5.0 hours there are

*Update* : I have

Am I good so far?

*My

A picture of my work

Seems like

Well, the forward reaction was:

#"Zn"(s) + "Cu"^(2+)(aq) -> "Zn"^(2+)(aq) + "Cu"(s)#

#E_(cell)^@ = "0.34 V" - (-"0.76 V")#

#= +ul"1.10 V"#

The reaction is spontaneous in the forward direction, so if one allows it to proceed naturally, it will do so by consuming

After **how much charge** was involved in this forward reaction.

#"3.0 C"/cancel"s" xx (60 cancel"s")/(cancel"1 min") xx (60 cancel"min")/(cancel"1 hr") xx 5.0 cancel"hr"#

#=# #"54000 C"#

Since according to the Faraday constant,

#54000 cancel"C" xx ("1 mol e"^(-))/(96485 cancel"C")#

#= "0.5597 mols e"^(-)#

Or, since **one atom** of **for every two** electrons to reduce it to

#0.5597 cancel("mols e"^(-)) xx ("1 mol Cu"^(2+))/(2cancel("mols e"^(-))#

#= "0.280 mols Cu"^(2+)#

were consumed in this

Or, since this is in **consumed**. This gives us an idea of how far we have gone towards equilibrium.

#["Cu"^(2+)]_"currently" = ["Cu"^(2+)]_0 - "0.280 M"#

#= "1.90 M" - "0.280 M" = "1.62 M Cu"^(2+)#

#["Zn"^(2+)]_"currently" = ["Zn"^(2+)]_0 + "0.280 M"#

#= "0.100 M" + "0.280 M" = "0.380 M Zn"^(2+)#

The **new**

#Q_c' = "0.380 M Zn"^(2+)/("1.62 M Cu"^(2+))#

#= ul0.234567#

Now we can find **instead of** initially.

#color(blue)(E_(cell)("5.0 hr")) = E_(cell)^@ - (RT)/(nF)lnQ_c'#

#= "1.10 V" - ("8.314 V"cdotcancel"C"//cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/(((2 cancel("mol e"^(-)))/cancel"1 mol atoms") cdot 96485 cancel"C"//cancel("mol e"^(-)))ln(0.234567)#

#=# #color(blue)("1.12 V")#