Question

A strip of nickel metal is placed in a \[1\] molar solution \[Ni\left( {NO{}_3} \right){}_2\]and a trip of silver metal is placed in \[1\] molar of \[AgNO{}_3\]. An electrochemical cell is created when two solutions are connected by a salt bridge and two strips are connected by wires of a voltmeter.
A.Write the balanced equation for overall reaction occurring in the cell and calculate the cell potential.
B.Calculate the cell potential, E at \[25^\circ \] Celsius for the cell if the initial concentration if \[Ni\left( {NO{}_3} \right){}_2\] is \[0.100\] molar and initial concentration of \[AgNO{}_3\]is \[1.00\] molar
\[[E^\circ {}_{\dfrac{{Ni{}^{2 + }}}{{Ni}}} = - 0.25;E^\circ {}_{\dfrac{{Ag}}{{Ag}}} = 0.80V,Log10{}^{ - 1} = - 1]\].

Answer 1

Hint: First we will equate the ionization of nickel and silver to find the overall reaction and balance it. Then by using the formula we will find the\[E{}_{cell}\]of the overall reaction and then find the cell potential of both the given solution at the given concentration.

Formula used: \[E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ \]

Complete step by step answer:
Let us solve the part one of the question and balance the overall equation;
The ionization of nickel:
\[Ni{}_{(s)} \to Ni_{(aq)}^{2 + } + 2e{}^ - \]
The ionization of silver:
\[2Ag_{(aq)}^ + + 2{e^ - } \to 2Ag{}_{(s)}\]
Hence the total reaction will be
\[Ni{}_{(s)} + 2Ag_{(aq)}^ + \to Ni_{(aq)}^{2 + } + 2Ag{}_{(s)}\]
Let us now solve the cell potential of overall reaction by the formula given below;\[2.5V\]
$
  E_{cell}^\circ = 0.80 - ( - 0.25) \\
  E_{cell}^\circ = 1.05V \\
$
The cell potential is \[Ni\left( {NO{}_3} \right){}_2\] and \[AgNO{}_3\]
\[E{}_{cell} = E_{cell}^\circ - \dfrac{{0.0591}}{2}\log \dfrac{{[Ni{}^{2 + }]}}{{[Ag{}^ + ]{}^2}}\]
We will now substitute the morality of ions:
$
  E{}_{cell} = E_{cell}^\circ - \dfrac{{0.0591}}{2}\log \dfrac{{0.1}}{{(1){}^2}} \\
  E{}_{cell} = 1.05 - \dfrac{{0.0591}}{2} \times (\log 10{}^{ - 1}) \\
  E{}_{cell} = 1.05 - 0.295 \times ( - 1) \\
  E{}_{cell} = 1.05 + 0.0295 \\
  E{}_{cell} = 1.075V \\
$
Hence the cell potential is \[1.075V\]

Note: An electrochemical cell is used for generation of electrical energy from chemical reactions. These cells which produce electric current are called voltaic or galvanic cells.
A cell potential can be assumed through the use of potentials. Cell potential has a range of \[0 - 6\] volts. Cells using water based electrolyte limits cell potential which is less than \[2.5V\](volts) due to very high reactivity of powerful oxidising agents and reducing agents with water that helps produce a higher voltage. Cell potential depends on concentration of reactants and their cell type.