Question

Calculate cell potential at 298K
$Z{n_{\left( s \right)}}/Z{n^{2 + }}_{(0.6M)}//C{u^{2 + }}_{(0.3M)}/C{u_{(s)}}$
${E^o}Z{n^{2 + }}/Zn = - 0.76V$
${E^o}C{u^{2 + }}/Cu = 0.34V$

Answer 1

Recall the nernst equation. You also need to calculate the standard cell potential for the given redox reaction. Write the half-cell reactions to get the clear idea of oxidation and reduction reactions. Here in the given reaction, Zn is getting oxidised, so its reaction will be at anode while Cu is getting reduced, so its reaction will be at cathode. Also, electron change for the reaction is 2.

Complete step by step solution:
Given equation for a chemical reaction is:
$Z{n_{\left( s \right)}}/Z{n^{2 + }}_{(0.6M)}//C{u^{2 + }}_{(0.3M)}/C{u_{(s)}}$
Here, at anode, Zn is converted to $Z{n^{2 + }}$ and at cathode, $C{u^{2 + }}$ is converted to Cu.
Oxidation and reduction half-cell reactions are:
At anode (oxidation): $Zn \to Z{n^{2 + }} + 2{e^ - }$
At cathode (reduction): $C{u^{2 + }} + 2{e^ - } \to Cu$
${E^o}_{cathode} = {E^o}_{C{u^{2 + }}/Cu} = 0.34V$ (Given)
${E^o}_{anode} = {E^o}_{Zn/Z{n^{2 + }}} = + 0.76V$ (Given)
Also, there is electron change of 2 in the reaction.
Standard cell potential, $E_{cell}^o$ for the reaction is:
Standard cell potential is the sum of oxidation potential at anode and the reduction potential at cathode. So, we can write $E_{cell}^o$for the given cell as:
$E_{cell}^o = {E^o}_{anode} + {E^o}_{cathode}$
$E_{cell}^o = {E^o}_{Zn/Z{n^{2 + }}} + {E^o}_{C{u^{2 + }}/Cu}$
Now, the nernst equation formula for calculating cell potential, ${E_{cell}}$for the given cell is:
${E_{cell}} = {E^o}_{cell} - \dfrac{{0.059}}{n}\log \dfrac{{{\text{[Product]}}}}{{{\text{[reactant]}}}}$
In the given reaction i.e, $Z{n_{\left( s \right)}}/Z{n^{2 + }}_{(0.6M)}//C{u^{2 + }}_{(0.3M)}/C{u_{(s)}}$
Electron change (n) = 2, on product side, concentration $Z{n^{2 + }}$ ion is 0.6M, on reactant side, concentration of $C{u^{2 + }}$ ion is 0.3 M. Thus,
${E_{cell}} = 1.10 - \dfrac{{0.059}}{2}\log \dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}$
${E_{cell}} = 1.10 - \dfrac{{0.059}}{2}\log \dfrac{{0.6}}{{0.3}}$
${E_{cell}} = 1.09V$

Hence, cell potential at 298K for the given cell is 1.09 V. Thus, this is the required answer.

Note: Actual nernst equation is:
${E_{cell}} = {E^o}_{cell} - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{{\text{[Product]}}}}{{{\text{[reactant]}}}}$
where, R is the gas constant i.e., $8.314J{K^{ - 1}}mo{l^{ - 1}}$, T is the temperature in Kelvin, and F is Faraday constant whose approximate value is 96500. At 298 K, putting standard values in the nernst equation, we get value of $\dfrac{{2.303RT}}{{nF}}$ as $\dfrac{{0.059}}{n}$ and equation thus become:
${E_{cell}} = {E^o}_{cell} - \dfrac{{0.059}}{n}\log \dfrac{{{\text{[Product]}}}}{{{\text{[reactant]}}}}$