Question

Calculate the minimum mass of$NaOH$ required to be added in$RHS$ to consume all the ${H^ + }$present in $RHS$of the cell of $EMF$ $ + 0.701$ volt at ${25^0}C$ before its use.
Also report the $EMF$ of the cell after addition of $NaOH$.
$Zn|\mathop {Z{n^{2 + }}}\limits_{0.1M} ||\mathop {HCl}\limits_{1litre} |\mathop {Pt({H_{2g}})}\limits_{1atm} ;$$E_{Zn/Z{n^{2 + }}}^0 = 0.760V.$

Answer 1

Hint: The voltage or electric potential difference across the terminals of a cell when no current is drawn from it. The electromotive force ($EMF$) is the sum of the electric potential differences produced by a separation of charges (electrons or ions) that can occur at each phase boundary (or interface) in the cell.

Complete Step by step answer: According to the give data of the question, the cell reaction will be:
 $Zn + 2{H^ + } \to Z{n^{2 + }} + {H_2}$
Applying Nernst equation for the above cell reaction:
 ${E_{cell}} = {E^0} - \dfrac{{0.0591}}{2}\log \dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}}$
 Given data of the cell are:${E_{cell}} = 0.701$ and${E^0} = 0.760$
Substituting the required values in the nernst equation stated above: $0.701 = 0.760 - \dfrac{{0.0591}}{2}\log \dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}}$
 So, $\log \dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}} = (0.701 - 0.760) \times \dfrac{2}{{0.0591}}$
$\log \dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}} = \dfrac{{0.0591 \times 2}}{{0.0591}} = 2$.
 So, $\dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}} = {10^2}$ = ${\left[ {{H^ + }} \right]^2} = \dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{10}^2}}}$. Now substitute the concentration of zinc ion $\left[ {Z{n^{2 + }}} \right] = 0.1$ in this equation.
We get, ${\left[ {{H^ + }} \right]^2} = \dfrac{{0.1}}{{{{10}^2}}} = {10^{ - 3}}$
Hence, $\left[ {{H^ + }} \right] = 0.0316mol{L^{ - 1}}$
Thus, 0.0316 mol/litre of $NaOH$ is required to neutralize${H^ + }$ ions.
So, mass of the $NaOH$ required will be $ = 0.0316 \times Molar\; mass;of\; NaOH$
Molar mass of $NaOH$ = 40
So, the required mass of $NaOH$ required $ = 0.0316 \times 40 = 1.264g$
Hence, the minimum mass of $NaOH$ required to be added in $RHS$ to consume all the ${H^ + }$present in $RHS$of the cell of $EMF$ $ + 0.701$ volt at ${25^0}C$ before its use is 1.264g.
Now come to the solution second part of the question:
After the addition of $NaOH$, the solution becomes neutral, the concentration of ${H^ + }$ ions in cathodic solution becomes ${10^{ - 7}}$ . Again applying Nernst equation,
${E_{cell}} = {E^0} - \dfrac{{0.0591}}{2}\log \dfrac{{\left[ {Z{n^{2 + }}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}}$
Substituting $\left[ {{H^ + }} \right] = {10^{ - 7}}$ and ${E_{cell}} = 0.701$ in the Nernst equation: ${E_{cell}} = 0.760 - \dfrac{{0.0591}}{2}\log \dfrac{{0.1}}{{{{\left( {{{10}^{ - 7}}} \right)}^2}}} = 0.3759volt$
${E_{cell}} = 0.3759volt$

Hence the $EMF$ after the addition of $NaOH$ is 0.3759 volt.

Note: The cell potential or $EMF$ of the electrochemical cell can be calculated by taking the values of electrode potentials of the two half – cells. There are usually three methods that can be used for the calculation: By taking into account the oxidation potential of anode and reduction potential of cathode, by considering the reduction potentials of both electrodes, by taking the oxidation potentials of both electrodes.