Question

A voltaic cell is set up at \[{{25}^{\circ }}C\]with the following half cells:
\[Al|A{{l}^{3+}}\left( 0.001M \right)\]and \[Ni|N{{i}^{2+}}\left( 0.50M \right)\]
Write an equation for the equation that occurs when the cell generates an electric current and determine the cell potential.
\[{{E}_{N{{i}^{2+}}/Ni}}=-0.25\]and \[{{E}^{\circ }}_{A{{l}^{3+}}/Al}=-1.66V\left( \log \left( 8\times {{10}^{-6}} \right)=-0.54 \right)\]

Answer 1

Hint: A voltaic cell is an electrochemical cell that utilizes a compound response to deliver electrical energy. The significant pieces of a voltaic cell: The anode is a terminal where oxidation happens. The cathode is an anode where decrease happens.

Complete step by step answer:
The cell potential, Ecell , is the proportion of the expected distinction between two half cells in an electrochemical cell. The potential contrast is brought about by the capacity of electrons to spill out of one half cell to the next. Electrons can move between terminals on the grounds that the synthetic response is a redox response.
An electric flow is a progression of electric charge in a circuit. All the more explicitly, the electric flow is the pace of charge stream past a given point in an electric circuit. The charge can be adversely charged electrons or positive charge transporters including protons, positive particles or openings.
Voltaic cell is a power creating cell which utilizes substance response. These cells utilize anode and cathode.
In this case, the aluminum terminal is anode and nickel cathode is the cathode.
In Aluminum cathode, Oxidation response happens.
\[2Al\to 2A{{l}^{3+}}+6e^-.....{{E}^{\circ }}=1.66V\]
In the nickel electrode, Reduction reaction takes place:
\[3N{{i}^{2+}}+6e^-\to 3Ni.....{{E}^{\circ }}=-0.25V\]
Balancing the overall reaction,
\[2Al+3N{{i}^{2+}}\to 2A{{l}^{3+}}+3Ni\]
Thus, the formula is
\[
 E_{cell}^{o}={{E}^{\circ }}cathode-{{E}^{\circ }}anode \\
 E_{cell}^{o}=-0.25+1.66 \\
 \Rightarrow E_{cell}^{o}=1.41V \\
\]
By using the nernst equation,
\[{{E}_{cell}}=E_{cell}^{o}-\dfrac{0.059}{6}\log \dfrac{{{\left[ A{{l}^{3+}} \right]}^{2}}}{{{\left[ N{{i}^{2+}} \right]}^{3}}}\]
\[
   =E_{cell}^{o}-\dfrac{0.059}{6}\log \dfrac{{{\left[ 0.001 \right]}^{2}}}{{{\left[ 0.5 \right]}^{3}}} \\
  =E_{cell}^{o}-0.00983\log \dfrac{{{10}^{-6}}}{0.125} \\
  =E_{cell}^{o}-0.00983\log 8\times {{10}^{-6}} \\
  =E_{cell}^{o}-0.008877+0.0598 \\
  =1.41-0.008877+0.0598 \\
  \Rightarrow {{E}_{cell}}=1.46011\,V
\]

Hence the answer is \[1.46011\,V\].

Note:
In electrochemistry, the Nernst condition is a condition that relates the decrease capability of an electrochemical response (half-cell or full cell response) to the standard anode potential, temperature, and exercises (regularly approximated by groupings) of the substance species going through decrease and oxidation .