Question

For the following electrochemical cell at 298K
$Pt(s)I{H_2}(g,1bar)I{H^ + }(aq,1M)\parallel {M^{4 + }}(aq),{M^{2 + }}(aq)IPt(s)$
${E_{cell}} = 0.092V\,when\,\dfrac{{[{M^{2 + }}(aq)]}}{{[{M^{4 + }}(aq)]}} = {10^x}$
Given: ${E^\circ}_{M^{4+}/M^{2+}} $ $= 0.151V; 2.303\dfrac{{RT}}{F} = 0.059{V^{}}$
The value of x is :
A. -2
B. -1
C. 1
D. 2

Answer 1

Hint:The overall cell potential can be calculated by using the equation $E{^\circ _{cell}} = {E_{red}}^\circ - {E_{oxid}}^\circ $ . Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction.

Complete answer:
An electrolytic cell has three component parts: an electrolyte and two electrodes (a cathode and an anode). The electrolyte is usually a solution of water or other solvents in which ions are dissolved. In a full electrochemical cell, species from one half-cell lose electrons (oxidation) to their electrode while species from the other half-cell gain electrons (reduction) from their electrode.A cell's standard state potential is the potential of the cell under standard state conditions, which is approximated with concentrations of 1 mole per liter (1 M) and pressures of 1 atmosphere at $25^\circ C$. Electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs. The anode is negative and cathode is the positive electrode.
In the above stated statement:
Anode: ${H_2}(g) \Leftrightarrow 2{H^ + } + 2{e^ - }$
Cathode: ${M^{4 + }}(aq) + 2{e^ - } \Leftrightarrow {M^{2 + }}(aq)$
Net cell reaction: ${H_2}(g) + {M^{4 + }}(aq) \Leftrightarrow {H_2}(g) + {M^{4 + }}(aq)$
Now
$E{^\circ _{cell}} = {E_{red}}^\circ - {E_{oxid}}^\circ $
${E_{cell}} = (E{^\circ _{{M^{4 +} /{M^{2+}}}}} - E{^\circ _{H^+ }}_{/{H_2}})$
Given:
${E_{cell}} = 0.092V\,when\,\dfrac{{[{M^{2 + }}(aq)]}}{{[{M^{4 + }}(aq)]}} = {10^x}$
$\Rightarrow {E_{cell}} = E{^\circ _{cell}} - \dfrac{{0.059}}{2}{\log _{10}}\dfrac{{[{M^{2 + }}]{{[{H^ + }]}^2}}}{{[{M^{4 + }}]P[{H_2}]}}$
$\Rightarrow 0.092 = 0.151 - \dfrac{{0.059}}{2}{\log _{10}}{10^x}$
$\Rightarrow x = 2$

Therefore the answer is 2. Hence the correct option is option (D).

Note:
The standard cell potential for a redox reaction ($E{^\circ _{CELL}}$) is a measure of the tendency of reactants in their standard states to form products in their standard states. The cell potential depends on the concentration of the reactants, as well as their type. As the cell is discharged, the concentration of the reactants decreases and the cell potential also decreases.