Question

The standard electrode potential of \[{{Cu/C}}{{{u}}^{{{ + 2}}}}\] is \[{{ - 0}}{{.34}}\] Volt. At what concentration of \[{{C}}{{{u}}^{{{ + 2}}}}\] ions, electrode potential will be zero?
A. \[{{1}}{{.98 \times 1}}{{{0}}^{{{ - 12}}}}\]M
B. \[{{2}}{{.98 \times 1}}{{{0}}^{{{ - 12}}}}\]M
C. \[{{1}}{{.49 \times 1}}{{{0}}^{{{ - 12}}}}\]M
D. \[{{5}}{{.96 \times 1}}{{{0}}^{{{ - 12}}}}\]M

Answer 1

Hint: Standard electrode potential can be obtained by taking the difference of standard potential of cathode and anode. Cathode is the right electrode and anode is the right electrode.

Complete step by step answer:
Oxidation is the loss of electrons and reduction is the gain of electrons. In a galvanic cell, the oxidation takes place at anode and reduction takes place at cathode. The two portions of the cell are called half cells or redox couples. Standard electrode potential is the electrode potential when the concentration of all the species involved is unit or it’s the potential at standard conditions like one molar concentration of all the ions, pressure is one bar and temperature is at \[{{298 K}}{{.}}\]
It is denoted as \[{{{E}}^{{0}}}_{{{cell}}}\].
The cell potential of an electrochemical cell at any given temperature, pressure, and reactant concentration is calculated using Nernst equation.
\[{{{E}}_{{{cell}}}}{{ = }}{{{E}}_{{0}}}{{ -2}}{{.303 }}\left[ {\dfrac{{{{RT}}}}{{{{nF}}}}} \right]{{\;log }}\left[ {\dfrac{{{{\left[ {{M}} \right]}^{{n}}}}}{{{{[}}{{{M}}^{{{n + }}}}{{]}}}}} \right]\]\[{{ = }}{{{E}}_{{0}}}{{ - }}\dfrac{{{{0}}{{.059}}}}{{{n}}}{{log }}\left[ {\dfrac{{{{\left[ {{M}} \right]}^{}}}}{{{{[}}{{{M}}^{{{n + }}}}{{]}}}}} \right]\]
Given that, \[{{{E}}^{{0}}}_{{{cell}}}\,{{ = - 0}}{{.34V}}\]and, \[{{{E}}_{{{cell}}}}\,\,{{ = }}\,\,{{0}}\]
The emf of electrode is given by Nernst equation,
We have to find the concentration of \[{{C}}{{{u}}^{{{2 + }}}}\],\[\left[ {{{Cu}}} \right] = \,1\](solid) and \[{{n}} = \,2\]
\[{{{E}}_{{{cell}}}}{{ = - 0}}{{.34 - }}\dfrac{{{{0}}{{.059}}}}{2}{{log }}\left[ {\dfrac{1}{{{{[C}}{{{u}}^{{{2 + }}}}{{]}}}}} \right]\],
\[0 = \, - {{0}}{{.34 + }}\dfrac{{{{0}}{{.059}}}}{2}{{log}}\left[ {{{C}}{{{u}}^{{{2 + }}}}} \right]\]

${{log}}\left[ {{{C}}{{{u}}^{{{2 + }}}}} \right]\,\, = \,\,\dfrac{{ - 0.34\,}}{{0.0295}}\,\, = \,\, - 11.52 \\
\left[ {{{C}}{{{u}}^{{{2 + }}}}} \right]\,\,\, \approx \,\,2.98\, \times \,\,{10^{ - 12}} \\$

Hence the answer is option B.

Note: The galvanic cell converts chemical energy into electrical energy., by means of redox reaction. It is also known as Daniel cell.
The standard electrode is measured by comparing its potential with respect to standard hydrogen electrode (SHE). The arrangement of electrodes in contact with electrolyte solutions based on their standard reduction potential or standard oxidation potentials is called the electrochemical series.