Question

Refer to the following standard reduction half-cell potentials at 25∘C:

VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answer 1

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions:  1 M , 1 atm).

Nernst equation at 25ºC is:

$E = E^{0} - [(\frac{0.0592}{n} · log Q)]$

where

E: Cell potential (non standard conditions)

$E^{0}$ = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

$Q = \frac{C^{c} * D^{d} }{A^{a}*B^{b} }$

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: $E^{0}$, n and Q

Let´s  calcule potential in nomal conditions  ($E^{0}$):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions:                               Eo (v):

$(VO_{2})^{2+}$ + e- --> $(VO_{2})^{+}$                -0.23

$Ni --> Ni^{2+}$  + 2 e-                   +0.99    

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions:                                                                Eo (v):

2 * [ $(VO_{2})^{2+}$ + e- --> $(VO_{2})^{+}$]                                          -0.23

1 * [$Ni --> Ni^{2+}$  + 2 e-     ]                                         +0.99  

-------------------------------------------------------------------         -------------

2 $(VO_{2})^{2+}$ + 2e- + Ni  --> 2$(VO_{2})^{+}$ + $Ni^{2+}$ + 2e-              0.76 v  

               

Then global balanced chemical reaction is:

2 $(VO_{2})^{2+}$ + Ni  --> 2$(VO_{2})^{+}$ + $Ni^{2+}$      

 and the potential in nomal conditions is:

$E^{0}$ = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

$Q = \frac{[(VO_{2}+]^{2}* [Ni^{2+} }{[VO_{2+}]^{2} * Ni }]$

From the exercise we know:

$[VO_{2} ^{2+}] = 2.5 M$

$[VO_{2}+] = 0.083 M$

$[Ni^{2+}] = 2.5 M$

$[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.$

$Q = \frac{(2.5)^{2}* 2.5 }{(0.083)^{2} * 1} = 2,268.11$

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

$E = E^{0} - (\frac{0.0592}{n} - log Q)\\ \\E = 0.76 - (\frac{0.0592}{2}-log (2,268.11) \\E = 0.76 - (0.0296 - 3.36)\\E = 4.09 v$

Cell potential under non standard concentration is 4.09 v