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Question: At \( {25^ \circ } \) ,Nernst equation will be \( A){E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0...

At 25{25^ \circ } ,Nernst equation will be
A)Ecell=Ecell0.0591nlog[ion]RHS[ion]LHSA){E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0591}}{n}\log \dfrac{{{{[ion]}_{RHS}}}}{{{{[ion]}_{LHS}}}}
B)Ecell=Ecell0.0591nlog[M]RHS[M]LHSB){E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0591}}{n}\log \dfrac{{{{[M]}_{RHS}}}}{{{{[M]}_{LHS}}}}
C)Ecell=Ecell+0.0591nlog[ion]RHS[ion]LHSC){E_{cell}} = {E^ \circ }_{cell} + \dfrac{{0.0591}}{n}\log \dfrac{{{{[ion]}_{RHS}}}}{{{{[ion]}_{LHS}}}}
D)Ecell=Ecell0.0591nlog[ion]LHS[ion]RHSD){E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0591}}{n}\log \dfrac{{{{[ion]}_{LHS}}}}{{{{[ion]}_{RHS}}}}

Explanation

Solution

Hint : To calculate the nernst equation at 25C{25^ \circ }C we will put the values for faraday’s constant and ideal gas constant at the given temperature. The value of Faraday's constant is 96485Cmol196485Cmo{l^{ - 1}} and value of ideal gas constant is 8.314.8.314.

Complete Step By Step Answer:
The Nernst equation is a state that relates the decrease potential of an electrochemical reaction to the standard cathode potential, temperature, and other activities of the compound species going through decrease and oxidation. It was named after Walther Nernst, a German actual scientist who defined the equation.
The Nernst equation has a physiological application when used to ascertain the potential of a particle of charge across a layer. This potential is resolved utilizing the grouping of the particle both inside and outside the cell:
= =E=RTzFln[ionoutsidecell][ioninsidecell]= E = \dfrac{{RT}}{{zF}}\ln \dfrac{{[ion outside cell]}}{{[ion inside cell]}} .
Where RR is ideal gas constant,
TT is temperature in kelvins,
FF is faraday’s constant,
zz is charge
At T=25CT = {25^ \circ }C ,substituting the value of F=96485Cmol1F = 96485Cmo{l^{ - 1}} and R=8.314R = 8.314 .
The equation we get,
=Ecell=Ecell0.0591nlnlogQ= {E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0591}}{n}\ln \log Q
Where Q=ionRHSionLHSQ = \dfrac{{io{n_{RHS}}}}{{io{n_{LHS}}}} .
=Ecell=Ecell0.0591nlogionRHSionLHS= {E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0591}}{n}\log \dfrac{{io{n_{RHS}}}}{{io{n_{LHS}}}}
So, the correct option is A)A) .

Additional Information:
The limitation of Nernst equation is that the movement of a particle in an exceptionally weakened arrangement is near infinity and can, along these lines, be expressed as far as the ion concentration. In any case, for arrangements having high concentration, the ion concentration isn't equivalent to the ion movement. To utilize the Nernst condition in such cases, experimental estimations should be led to get the genuine activity of the particle.
Another limitation of this condition is that it can't be utilized to measure cell potential when there is a current moving through the electrode. This is on the grounds that the progression of current influences the action of the particles on the outside of the cathode. Likewise, extra factors like resistive force and over potential should be viewed as when there is a current flowing through the anode.

Note :
The Nernst condition can likewise be utilized to decide the complete voltage for a full electrochemical cell. The Nernst condition gives a formula that relates the numerical estimations of the concentration gradient to the electric gradient that adjusts it.