# Electrochemistry question? (Voltaic Cell)

##  A: Which electrode is the cathode in the reaction above? B: Give the half reaction occurring at each of the electrodes below. Make sure it shows the direction the reaction actually proceeds in the cell above. C: Write a balanced chemical equation describing the total reaction associated with the electrochemical cell. D: What is the EMF of the cell above if it is operated under standard state conditions?

Apr 10, 2017

I will break it down for you so you can understand how to look at these types of questions.

#### Explanation:

This is a Galvanic cell. Galvanic cells are spontaneous meaning they have a $- \Delta {G}^{\circ}$. A $- \Delta {G}^{\circ}$ just means the reaction occurs without the input of any outside energy source, under standard conditions that is. The standard cell potential of the cell, ${E}^{\circ}$, is always positive in a galvanic cell. Basically, Gibbs and Standard Cell Potential have opposite signs.

You have solid $\text{Cu}$ and $\text{Zn}$ in their respective half cells. Looking at the standard reduction potentials provided, the more positive value means the more likely that species is going to be reduced. A quick look at the table will show you that ${\text{Cu}}^{+ 2}$ ions will readily reduce to form solid $\text{Cu}$ than ${\text{Zn}}^{+ 2}$ ions will to become $\text{Zn}$ solid.

color(white)(aaaa)color(magenta)["Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s) color(white)(aaaaa)"E"^@("red") = +0.34] $\left(\text{more positive red. potential means reduction will occur}\right)$

color(white)(aaaa)color(green)["Zn"^(+2)(aq) + 2e^(-)->"Zn"(s) color(white)(aaaaa)"E"^@("red") = -0.76] $\left(\text{less positive red. potential means oxidation occurs}\right)$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$
$- - - - - - - - - - - - - - - - - - -$

$\textcolor{b l u e}{\text{A: Which electrode is the cathode in the reaction above?}}$

• Since we know that${\text{Cu}}^{+ 2}$ ions are more likely to be reduced to become $\text{Cu} \left(s\right)$, the $\textcolor{red}{\text{Cu electrode is labeled as the cathode}}$ while the $\textcolor{red}{\text{Zn electrode is labeled as the anode}}$, where oxidation occurs. (remember the memory aid AN OX, RED CAT)

$\textcolor{b l u e}{\text{B: Give the half reaction occurring at each of the electrodes below.}}$

$\textcolor{b l u e}{\text{Make sure it shows the direction the reaction actually proceeds in the}}$
$\textcolor{b l u e}{\text{cell above.}}$

• If reduction is occurring at the $\text{Cu}$ cathode (where we know the solution of ${\text{Cu(NO3)}}_{2}$ ionizes to give off ${\text{Cu}}^{+ 2}$ cations and ${\text{NO}}_{3}^{- 1}$ anions), this means ${\text{Cu}}^{+ 2}$ ions will gain $2 {e}^{-}$ and will deposit onto the $\text{Cu}$ electrode. The half reaction is the following for the left half cell:

color(white)(aaaaaaaaaaaa)color(red)["Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s)

$\textcolor{w h i t e}{a a a a a a a a a a a a}$

• In the right half cell, the $\text{Zn}$ electrode (which is the anode) will lose $2 {e}^{-}$ and ${\text{Zn}}^{+ 2}$ ions will fall out into the solution which already has ${\text{Zn}}^{+ 2}$ cations and ${\text{NO}}_{3}^{- 1}$ ions from the zinc nitrate solution. The half reaction in the right cell is the following:

color(white)(aaaaaaaaaaa)color(red)["Zn"(s)-> "Zn"^(+2)(aq) + 2e^(-)

$\textcolor{b l u e}{\text{C: Write a balanced chemical equation describing the total reaction of the}}$
$\textcolor{b l u e}{\text{electrochemical cell.}}$

• If we combine the two half reactions, we get

$\text{Cu"^(+2)(aq) + cancel(2e^(-)) ->"Cu} \left(s\right)$
${\text{Zn"(s)-> "Zn}}^{+ 2} \left(a q\right) + \cancel{2 {e}^{-}}$
$- - - - - - - - - - - - - -$
color(red)["Cu"^(+2)(aq) + "Zn"(s)->"Cu"(s) + "Zn"^(+2)(aq)

$\textcolor{b l u e}{\text{D: What is the EMF of the cell above(under standard state conditions)}}$

• Okay. We have established that at the $\text{Cu}$ half cell, reduction will occur and at the $\text{Zn}$ half cell. oxidation will occur. To find the EMF of the cell, the following equation is going to be used.

color(white)(aaaaaaa)"E"^@("cell") = "E"^@("red") + "E"^@("oxidation")

Since we were provided "E"^@("red") potentials, we have to reverse the reaction to show the oxidation of $\text{Zn} \left(s\right)$ as well as reversing the value of the standard reduction potential to get the standard oxidation potential.

"Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s)color(white)(aaaaa)"E"^@("red") = +0.34

"Zn"(s)->"Zn"^(+2)(aq) + 2e^(-)color(white)(aaaaa)"E"^@("oxidation") = +0.76
$\textcolor{w h i t e}{a a a a a}$
- Use the equation to plug in and solve:

color(white)(aaaaa)"E"^@("cell") = "E"^@("red") + "E"^@("oxidation")

$\textcolor{w h i t e}{a a a a a} \textcolor{red}{\text{E"^@("cell}} = \left(+ 0.34\right) + \left(+ 0.76\right) = + 1.1$