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

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.