| The emf of a Daniell cell at 298 K is\[{{E}_{1}}\] [AIPMT 2003] |
| \[ZN\,\left| \begin{align} & ZnS{{O}_{4}} \\ & (0.01\,M) \\ \end{align} \right|\,\left| \begin{align} & CuS{{O}_{4}} \\ & (1.0\,M) \\ \end{align} \right|Cu\] |
| When the concentration of \[ZnS{{O}_{4}}\] is 1.0 M and that of \[CuS{{O}_{4}}\] is 0.01 M, the emf changed to \[{{E}_{2}}.\]What is the relationship between \[{{E}_{1}}\] and \[{{E}_{2}}\]? |
A) \[{{E}_{1}}={{E}_{2}}\]
B) \[{{E}_{2}}=0\,\ne {{E}_{1}}\]
C) \[{{E}_{1}}>E{{ & }_{2}}\]
D) \[{{E}_{1}}<{{E}_{2}}\]
Correct Answer: C
Solution :
| [c] For Daniell cell |
| \[Zn\left| \begin{align} & ZnS{{O}_{4}} \\ & 0.01\,M \\ \end{align} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\left| \begin{align} & CuS{{O}_{4}} \\ & 1.0\,M \\ \end{align} \right|\,\,Cu\] |
| Cell reaction is |
| \[Zn(s)+C{{u}^{2+}}(aq)\rightleftharpoons Cu(s)+Z{{n}^{2+}}(aq)\] |
| For above cell |
| \[{{E}_{1}}={{E}^{o}}_{cell}-\frac{0.0591}{n}{{\log }_{10}}\frac{[Z{{n}^{2+}}]}{[Cu]}\] |
| \[{{E}_{1}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{0.01}{1.0}\] |
| \[{{E}_{1}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{1}{100}\] |
| \[=E_{cell}^{o}+0.0591\,\,\,{{\log }_{10}}\,10\] |
| \[=E_{cell}^{o}+0.0591\] ....(i) |
| When the concentration of \[Z{{n}^{2+}}\] is 1.0 M and concentration of \[C{{u}^{2+}}\,\] is 0.01 M |
| \[{{E}_{2}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{1}{0.01}\] |
| \[=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}{{10}^{2}}\] |
| \[=E_{cell}^{o}-0.0591\] (ii) |
| From Eqs. (i) and (ii) \[{{E}_{1}}>{{E}_{2}}\] |
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