Question

What is the cell potential (standard emf, ${{{E}}^{{0}}}$) for the reaction below?
$\left[ {{{{E}}^{{0}}}_{\left( {_{{{F}}{{{e}}^{{{2 + }}}}\left( {{{aq}}} \right){{/Fe}}}} \right)}{{ = 0}}{{.44 V and }}{{{E}}^{{0}}}{{{ }}_{\left( {_{{{{O}}_{{2}}}\left( {{g}} \right){{/Fe}}}} \right)}}{{ = + 0}}{{.4 V}}} \right]$
${{2Fe}}\left( {{s}} \right){{ + }}\,{{{O}}_{{2}}}\left( {{g}} \right){{ + }}\,{{{H}}_{{2}}}{{O}}\left( {{l}} \right)\,\, \to {{2F}}{{{e}}^{{{2 + }}}}\,{{ + }}\,{{4O}}{{{H}}^{{ - }}}$
A. ${{{E}}^{{0}}}{{ cell = - 0}}{{.48 V }}$
B. ${{{E}}^{{0}}}{{ cell = - 0}}{{.04V }}$
C. ${{{E}}^{{0}}}{{ cell = + 0}}{{.84V }}$
D. ${{{E}}^{{0}}}{{ cell = + 1}}{{.28V }}$

Answer 1

Standard electrode potential is the potential at standard conditions. The standard EMF of a reaction is obtained by taking the difference of standard potential of cathode and anode. Cathode is the right electrode and anode is the right electrode. Here, ${\text{Fe}}\,$ is the anode it undergoes oxidation and ${{\text{O}}_{\text{2}}}$ is the cathode, it undergoes reduction.

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.
When two-electrodes are dipped in solution containing their electrolyte solution, the more reactive electrodes undergo oxidation whereas the less reactive electrodes undergo reduction also there is a tendency of metal atoms to go into the solution.
A potential difference develops between the electrode and electrolyte is called electrode potential. The cell potential is the difference between the electrode potential of cathode and anode; it is called the emf (electromotive force) of the cell when no current is drawn through the cell.
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 ${{{E}}^{{0}}}_{{{cell}}}$
${{{E}}^{{0}}}_{{{cell}}}{{ = }}{{{E}}^{{0}}}_{{{cathode}}}\,{{ - }}\,{{{E}}^{{0}}}_{{{anode}}}$
Or
${{{E}}^{{0}}}_{{{cell}}}{{ = }}{{{E}}^{{0}}}_{{{right}}}\,{{ - }}\,{{{E}}^{{0}}}_{{{left}}}$
A negative ${{{E}}^{{0}}}_{{{cell}}}$ means the redox couple is strong reducing agent that hydrogen and a positive ${{{E}}^{{0}}}_{{{cell}}}$ means the redox couple is strong oxidising agent than hydrogen.
Consider the reaction,
${{2Fe}}\left( {{s}} \right){{ + }}\,{{{O}}_{{2}}}\left( {{g}} \right){{ + }}\,{{{H}}_{{2}}}{{O}}\left( {{l}} \right)\,\, \to {{2F}}{{{e}}^{{{2 + }}}}\,{{ + }}\,{{4O}}{{{H}}^{{ - }}}$
Here ${{Fe}}\,$ undergoes oxidation and ${{{O}}_{{2}}}$ undergoes reduction. ${{Fe}}\,$ is the anode and ${{{O}}_{{2}}}$ is the cathode? Therefore, Standard electrode potential ${{{E}}^{{0}}}_{{{cell}}}$ is,
{{{E}}^{{0}}}_{{{cell}}}{{ = }}{{{E}}^{{0}}}_{{{cathode}}}\,{{ - }}\,{{{E}}^{{0}}}_{{{anode}}}$$$${{ = }}\left( {{{ 0}}{{.40}}} \right){{ - }}\left( {{{ - 0}}{{.44}}} \right)\,{{ = }}\,{{0}}{{.84V}}

The, correct answer is C.

Note: The standard electrode 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.
The galvanic cell converts chemical energy into electrical energy., by means of redox reaction. It is also known as Daniel cell.