Question

5. A strip of nickel metal is placed in a 1 molar solution of Ni(NO), and a strip of silver metal is
placed in a 1-molar solution of AgNO,. An electrochemical cell is created when the two solutions are
connected by a salt bridge and the two strips are connected by wires to a voltmeter.
(1) Write the balanced equation for the overall reaction occurring in the cell and calculate the cell
potential.
(ii) Calculate the cell potential, E, at 25°C for the cell if the initial concentration of Ni(NO2), is 0.100
molar and the initial concentration of AgNO, is 1.00 molar.
[E2/= -0.25 V; E = 0,80 V, log 10-4 = -1]​

Answer 1

(1) Balanced equation for the overall reaction:

Ionisation of nickel:

$\mathrm{Ni}_{(s)} \rightarrow \mathrm{Ni}^{2+}_{(a q)}+2 e^{-}$

Ionisation of silver:

$2 \mathrm{Ag}^{+}_{(a q)}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}_{(s)}$

Thus, the total reaction is:

$\mathrm{Ni}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow \mathrm{Ni}^{2+}_{(a q)}+2 \mathrm{Ag}_{(\mathrm{s})}$

Cell  potential for the overall reaction:

The cell potential is given by the formula:

$\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\text {cathode }}^{0}-\mathrm{E}_{\text {anode }}^{0}$

$\mathrm{E}^{0} \text { cell }=0.80-(-0.25)$

$\therefore \mathrm{E}^{0} \text { cell } = 1.05 \ V$

(2) Cell  potential:

The cell potential is Ni(NO₂) and AgNO

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{0}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$

On substituting the molarity of the ions, we get,

$\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{0}-\frac{0.0591}{2} \log \frac{0.1}{(1)^{2}}$

$\mathrm{E}_{\mathrm{cell}}=1.05-\frac{0.0591}{2} \times\left(\log 10^{-1}\right)$

$\mathrm{E}_{\mathrm{cell}}=1.05-0.295 \times(-1)$

Thus, the cell potential is:

$\mathrm{E}_{\mathrm{cell}}=1.05+0.0295$

$\therefore\mathrm{E}_{\mathrm{cell}}=1.0795 \ \mathrm{V}$