VO_2^+(aq) + 2H^+(aq) + e^– -> VO_2^+(aq) + H_2O, Eo = +1.00 V Cr_2O_72–(aq) + 14H+(aq) + 6e– -> 2Cr_3^+(aq) + 7H_2O, Eo = +1.33 V How do you write the balanced equation for the spontaneous process and determine its standard cell potential?

Jun 9, 2016

Here's how you can do it:

Explanation:

I am assuming we have vanadium(V) as $V {O}_{2}^{+}$ and vanadium(IV) as $V {O}^{2 +}$.

List the 1/2 equations in order least positive to most positive:

 " " "E"^@("V")

stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)

$V {O}_{2 \left(a q\right)}^{+} + 2 {H}_{\left(a q\right)}^{+} + e r i g h t \le f t h a r p \infty n s V {O}_{\left(a q\right)}^{2 +} + {H}_{2} {O}_{\left(l\right)} \text{ } + 1.00$

$C {r}_{2} {O}_{7 \left(a q\right)}^{2 -} + 14 {H}_{\left(a q\right)}^{+} + 6 e r i g h t \le f t h a r p \infty n s 2 C {r}_{\left(a q\right)}^{3 +} + 7 {H}_{2} {O}_{\left(l\right)} \text{ } + 1.33$

stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)

Note we use the $r i g h t \le f t h a r p \infty n s$ symbol to show that the 1/2 cells can go in either direction depending on what they are coupled with.

The 1/2 cell with the most +ve ${\text{E}}^{\circ}$ is the one which will take in the electrons.

From this we can see that the 2nd 1/2 cell will be driven left to right and the 1st 1/2 cell right to left in accordance with the arrows.

So the 2 half - reactions are:

$V {O}_{\left(a q\right)}^{2 +} + {H}_{2} {O}_{\left(l\right)} \rightarrow V {O}_{2 \left(a q\right)}^{+} + 2 {H}_{\left(a q\right)}^{+} + e \text{ } \textcolor{red}{\left(1\right)}$

$C {r}_{2} {O}_{7 \left(a q\right)}^{2 -} + 14 {H}_{\left(a q\right)}^{+} + 6 e \rightarrow 2 C {r}_{\left(a q\right)}^{3 +} + 7 {H}_{2} {O}_{\left(l\right)} \text{ } \textcolor{red}{\left(2\right)}$

To get the electrons to balance you can see that we need to multiply $\textcolor{red}{\left(1\right)}$ by $6$ and add this to equation $\textcolor{red}{\left(2\right)} \Rightarrow$

6VO_((aq))^(2+)+6H_2O_((l))+Cr_2O_(7(aq))^(2-)+cancel(14H_((aq))^(+))+cancel(6e)rarr6VO_(2(aq))^(+)+cancel(12H_((aq))^(+))+cancel(6e)+ 2Cr_((aq))^(3+)+7H_2O_((l)) "

This simplifies to:

6VO_((aq))^(2+)+6H_2O_((l))+Cr_2O_(7(aq))^(2-)+2H_((aq))^(+)rarr6VO_(2(aq))^(+)+ 2Cr_((aq))^(3+)+7H_2O_((l)) "

To find ${\text{E}}_{c e l l}^{\circ}$ you subtract the least +ve ${\text{E}}^{\circ}$ from the most +ve ${\text{E}}^{\circ}$ value.

This means that ${\text{E}}_{c e l l}^{\circ}$ always has a +ve value which must be the case for the reaction to be spontaneous under standard conditions.

So in this case:

$\text{E"_(cell)^@=1.33-1.0=0.33"V}$

The advantage of this convention is that you do not have to make assumptions about the direction of the 1/2 cells and manipulate their signs.