Electrochemistry
A galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy whereas, an electrolytic cell is a device which uses electrical energy to carry out the non-spontaneous chemical reactions. We can construct innumerable number of galvanic cells on the pattern of Daniell cell (Cu-Zn cell) by taking combinations of different half-cells. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution and in an accepted convention we keep the anode on the left and the cathode on the right side. The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell.
Find the emf of the given cell at 298 K, if concentration of aq. nickel chloride is 0.1 M and aq. silver nitrate is 0.01 M.
Ni(s)|Ni2+(aq.)||Ag+(aq.)|Ag(s)
Given \( E^{\circ}_{\mathrm{Ni^{2+}/Ni}} \) = −0.25 V, \( E^{\circ}_{\mathrm{Ag^{+}/Ag}} \) = +0.80 V
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.96 V
The standard emf of a cell holds only when every ion is at 1 M. At other concentrations the emf is found from the Nernst equation, which at 298 K is:
\( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{n}\log Q \)
Here n is the number of electrons transferred and Q is the reaction quotient. Pure solids (the metals) are left out of Q, and each concentration is raised to the power of its coefficient.
The nickel electrode is the anode (oxidation) and the silver electrode is the cathode (reduction):
\( \displaystyle \text{Anode: }\ce{Ni(s) -> Ni^2+(aq) + 2e^-} \)
\( \displaystyle \text{Cathode: }\ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \)
\( \displaystyle \text{Cell: }\ce{Ni(s) + 2Ag^+(aq) -> Ni^2+(aq) + 2Ag(s)},\quad n=2 \)
\( \displaystyle E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}=0.80-(-0.25)=1.05\ \mathrm{V} \)
\( \displaystyle Q=\frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^{2}}=\frac{0.1}{(0.01)^{2}}=\frac{10^{-1}}{10^{-4}}=10^{3} \)
\( \displaystyle E_{\text{cell}}=1.05-\frac{0.059}{2}\log10^{3} \)
\( \displaystyle E_{\text{cell}}=1.05-0.0295\times3 \)
\( \displaystyle E_{\text{cell}}=1.05-0.0885 \)
\( \displaystyle E_{\text{cell}}=0.9615\ \mathrm{V}\approx0.96\ \mathrm{V} \)
So the emf of the cell is 0.96 V.